SK Tuitions • Class 10 Mathematics

Some Applications of Trigonometry — Heights & Distances

A complete virtual teacher, concept revision module, diagram-based learning system, solved question bank, CBSE practice resource and HOTS/Olympiad training module.

95 Main Entries
30 MCQs + 15 two-mark + 15 three-mark + 20 four-mark + 10 five-mark + 5 case studies.

Diagram First
Learn to convert word problems into right triangles before selecting a ratio.

Progressive Difficulty
Foundation → CBSE/Exemplar → Competency/HOTS → Olympiad challenge.

1. Essential Trigonometry Revision

Applications of trigonometry are modelling problems. The geometry comes first; the ratio comes second. In a right triangle, relative to the chosen angle θ:

sin θ = PerpendicularHypotenuse

cos θ = BaseHypotenuse

tan θ = PerpendicularBase

cot θ = BasePerpendicular

Key strategy

In many height-and-distance problems, the vertical height and horizontal distance are the two relevant sides, so tan θ = vertical height / horizontal distance. Do not use tan blindly: first identify the sides known and required.

θ	0°	30°	45°	60°	90°
sin θ	0	1/2	1/√2	√3/2	1
cos θ	1	√3/2	1/√2	1/2	0
tan θ	0	1/√3	1	√3	Not defined
cot θ	Not defined	√3	1	1/√3	0

2. Line of Sight, Angle of Elevation & Angle of Depression

Line of sight is the straight line joining the observer’s eye to the object being viewed. A horizontal line through the eye is the reference line for elevation or depression.

Angle of elevation: observer looking upward from a horizontal eye-level line.

Angle of depression: observer above the object looking downward from a horizontal line.

Angle of elevation: the angle between the horizontal through the observer’s eye and the upward line of sight.

Angle of depression: the angle between the horizontal through the observer’s eye and the downward line of sight.

Common Mistake

The angle of elevation is located at the observer, not at the top of the tower. In a depression problem, first draw the horizontal through the observer; do not place the depression angle arbitrarily inside the lower triangle.

Why angle of depression = corresponding angle of elevation

The horizontal through the observer and the horizontal ground/sea level are parallel. The line of sight acts as a transversal, so the relevant alternate interior angles are equal.

3. How to Convert a Word Problem into a Trigonometric Diagram

1. Identify vertical object

2. Draw vertical

3. Draw horizontal ground

4. Mark observer

5. Draw line of sight

6. Mark angle

7. Mark data

Add the observer’s eye height if it is given.
Mark the right angle where the vertical object meets the horizontal ground or eye-level line.
Identify the exact right triangle you will use.
Only after that, choose sin, cos or tan.

Exam Tip

Mark the 90° angle explicitly. A correct diagram prevents most sign, distance and observer-height errors.

4. Basic Height-and-Distance Model

A student stands 20 m from the foot of a tower. The angle of elevation of the top is 45°. Let the height be h.

A 20 m horizontal distance and a 45° line of sight to the tower top.

Known sides: vertical height h and horizontal distance 20 m.

Use tan: tan 45° = h / 20.

Since tan 45° = 1, we get 1 = h / 20.

Therefore, h = 20 m.

5. Choosing the Correct Trigonometric Ratio

Height + horizontal distance

Usually: tan θ = P/B

Height + line of sight

Usually: sin θ = P/H

Horizontal distance + line of sight

Usually: cos θ = B/H

Choose the ratio containing what is known and what is required, while avoiding an unnecessary third side.

6. Observer’s Height

If an observer’s eye is 1.5 m above the ground, the trigonometric triangle often gives only the height of the object above eye level.

Observer-height model showing eye level, upper tower portion and total height.

Total height = height above eye level + observer’s eye height

Common Mistake

If the question asks for the total height, do not stop after finding the portion above the horizontal through the observer’s eye.

7. Objects Below the Observer: Cliffs, Lighthouses, Terraces & Bridges

When the observer looks downward, convert the angle of depression into the equal angle of elevation in the lower right triangle. Then apply the required trigonometric ratio.

tan θ = vertical heighthorizontal distance

Why this step?

The sea level or ground is horizontal and parallel to the horizontal through the observer. The line of sight is a transversal, so the two corresponding acute angles are equal.

8. Two Observation Points on the Same Side

Two observers on the same side of a tower, with the nearer observer seeing the larger angle.

Let the nearer horizontal distance be x, the separation of observers be d, and tower height be h.

From the nearer point: tan β = h/x. …(1)

From the farther point: tan α = h/(x + d). …(2)

Use the known values of α, β and d to eliminate x and find h.

Check the geometry: the nearer point should normally have the larger angle.

9. Two Observation Points on Opposite Sides

If the tower stands between two observers at horizontal distances x and y, then:

tan α = h/x and tan β = h/y

x + y = distance between the observers

Use the two trigonometric relations with the linear distance equation.

10. Moving Towards or Away from a Tower

An observer moves from a farther position to a nearer position while viewing the same tower.

For a fixed tower height, moving closer decreases horizontal distance and therefore increases the angle of elevation. Moving away does the reverse.

Competency idea

The height remains fixed. Write one trigonometric equation from each position, then use the distance travelled to connect the two horizontal distances.

11. Two Buildings / Top and Bottom of Another Object

Observer at one building seeing the top and foot of another building using elevation and depression.

Such questions usually contain two right triangles: one below eye level and one above eye level. First find the common horizontal distance from the depression triangle; then find the height difference from the elevation triangle.

Other building height = height below eye level + height above eye level

12. Flagpoles, Antennas and Objects on Buildings

A flagstaff on a building viewed from a point on the ground at two angles.

If a flagstaff of height x stands on a building of height h, form one equation to the top of the building and a second equation to the top of the flagstaff.

Total height = h + x

13. Shadow Problems

The sun’s ray joining the top of the object to the end of its shadow plays the role of the line of sight.

tan θ = heightshadow length

Shadow applications are useful enrichment and may also appear in competency-style school questions.

14. Exact Values and Rationalisation

For standard angles, keep exact surd values unless the question asks for a decimal approximation.

30/√3 = (30/√3) × (√3/√3) = 10√3

If required, use √3 ≈ 1.732 only at the final stage.

15. Diagram Reading Skills

Identify observer, horizontal, line of sight, perpendicular, base and hypotenuse.

Identify the angle of depression and its equal corresponding angle of elevation.

Perpendicular: the vertical height relative to the chosen angle.
Base: the horizontal separation.
Hypotenuse: the line of sight in the right triangle.
Never assume a diagram is drawn to scale; use only labelled information.

16. Applications of Trigonometry — Formula & Strategy Sheet

Basic ratios
sin θ = P/H
cos θ = B/H
tan θ = P/B

Most common model
tan θ = vertical height / horizontal distance

Standard values
tan 30° = 1/√3
tan 45° = 1
tan 60° = √3

Angle rule
angle of depression = corresponding angle of elevation

Observer height
total height = height above eye level + eye height

Two-position rule
write one equation from each right triangle, then combine.

The 7-Step Method

Read

Draw

Mark

Find right triangle

Known vs required

Choose ratio

Solve + units

17. Explore Heights and Distances

Horizontal distance: 20 m

Angle of elevation: 45°

Observer eye height: 0 m

Interactive explorer showing how height changes with horizontal distance, angle and observer height.

h = d tan θ

18. 10 Common Mistakes in Applications of Trigonometry

Drawing the angle of elevation at the object instead of the observer.
Confusing elevation with depression.
Forgetting that corresponding elevation and depression angles are equal.
Using the wrong trigonometric ratio.
Treating line-of-sight distance as horizontal distance.
Forgetting observer/eye height.
Using an incorrect standard-angle value.
Adding or subtracting two observation-point distances incorrectly.
Omitting units in the final answer.
Assuming the figure is drawn to scale.

Common ratio error

Incorrect: tan θ = horizontal distance / height. Correct for the usual tower model: tan θ = height / horizontal distance.

19. How to Score Full Marks in Applications of Trigonometry

Draw a neat labelled figure and show the right-angle mark.
Define the unknown height or distance.
Name the right triangle or clearly identify the sides being used.
Write the trigonometric ratio before substituting values.
Use exact standard values and rationalise where required.
For multiple positions, show equation (1), equation (2), then eliminate systematically.
Add observer height only when the final quantity requires it.
Interpret the numerical result in context and write correct units.
Box or clearly highlight the final answer.

Section A — 30 MCQs (30 × 1 Mark)

Question 1

1 Mark Level 1 — Foundation NCERT Type Line of Sight

Which statement best defines the line of sight?

The vertical distance between observer and object

The straight line joining the observer’s eye to the object being viewed

The horizontal distance from the observer to the object

The perpendicular dropped from the object to the ground

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

A line of sight directly joins the observer’s eye to the point being viewed.

The horizontal and vertical lines are auxiliary reference lines; they are not the line of sight.

Therefore, option B is correct.

Question 2

1 Mark Level 1 — Foundation NCERT Type Elevation

An observer standing on level ground looks at the top of a tower. Where is the angle of elevation measured?

At the top of the tower

At the foot of the tower

At the observer, between the horizontal and the upward line of sight

Between the tower and the ground at its foot

Angle of elevation is measured at the observer.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: C

The reference horizontal is drawn through the observer’s eye.

The upward line of sight meets this horizontal at the observer.

Hence the angle of elevation is at the observer: option C.

Question 3

1 Mark Level 1 — Foundation NCERT Type Depression

From the top of a lighthouse, the angle of depression of a boat is 35°. What is the corresponding angle of elevation of the lighthouse top from the boat?

35°

55°

145°

Cannot be determined

Equal depression and elevation angles formed by parallel horizontals.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: A

The horizontal through the lighthouse top is parallel to the horizontal sea level.

The line of sight is a transversal, so the corresponding alternate interior acute angles are equal.

Angle of elevation = 35°. Option A.

Question 4

1 Mark Level 1 — Foundation NCERT Type Standard Values

The value of tan 60° is:

1/√3

√3

2/√3

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: C

From the standard trigonometric table, tan 60° = √3.

Option C is correct.

Question 5

1 Mark Level 1 — Foundation NCERT Type Tower Height

A pole is viewed from a point 12 m from its foot at an angle of elevation 45°. Ignoring observer height, the height of the pole is:

6 m

12 m

12√3 m

24 m

A 45° line of sight from 12 m away.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Let the height be h.

tan 45° = h/12.

1 = h/12, so h = 12 m.

Option B.

Question 6

1 Mark Level 1 — Foundation NCERT Type Tower Height

A tree is 10 m horizontally from an observer. If the angle of elevation of its top is 30°, its height above the observer’s eye level is:

10√3 m

10/√3 m

5√3 m

20 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Let the required height be h.

tan 30° = h/10.

1/√3 = h/10 ⇒ h = 10/√3 = 10√3/3 m.

Option B represents the same exact value.

Question 7

1 Mark Level 1 — Foundation NCERT Type Horizontal Distance

A tower has height 15√3 m. From a point on level ground, its top is seen at 60°. The horizontal distance from the tower is:

5 m

15 m

15√3 m

45 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Let the distance be d.

tan 60° = 15√3/d.

√3 = 15√3/d ⇒ d = 15 m.

Option B.

Question 8

1 Mark Level 1 — Foundation NCERT Type Observer Height

A student’s eye is 1.5 m above the ground. From a point 20 m from a tower, the angle of elevation of its top is 45°. The tower’s total height is:

20 m

21.5 m

18.5 m

20.5 m

Observer eye height added to the height above eye level.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Height above eye level = h.

tan 45° = h/20 ⇒ h = 20 m.

Total height = 20 + 1.5 = 21.5 m.

Option B.

Question 9

1 Mark Level 1 — Foundation NCERT Type Ratio Selection

If a problem gives only a vertical height and a horizontal distance, which trigonometric ratio usually connects them most directly?

sin θ

cos θ

tan θ

sec θ

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: C

Relative to the observation angle, vertical height is the perpendicular and horizontal distance is the base.

tan θ = perpendicular/base.

Option C.

Question 10

1 Mark Level 1 — Foundation NCERT Type Ratio Selection

If the vertical height and the line-of-sight distance are the two relevant quantities, the most direct ratio is:

sin θ

cos θ

tan θ

cot θ

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: A

Vertical height is the perpendicular and line of sight is the hypotenuse.

sin θ = perpendicular/hypotenuse.

Option A.

Question 11

1 Mark Level 1 — Foundation NCERT Type Conceptual

For a fixed tower height on level ground, which observer sees the larger angle of elevation?

The observer farther away

The observer nearer the tower

Both see the same angle

It depends only on tower colour

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

For fixed height h, tan θ = h/d.

When d decreases, h/d increases; for acute angles, θ therefore increases.

The nearer observer sees the larger angle: option B.

Question 12

1 Mark Level 1 — Foundation NCERT Type Shadow

A vertical pole casts a shadow 8√3 m long when the angle of elevation of the sun is 30°. The pole’s height is:

8 m

8√3 m

24 m

4√3 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: A

Let height = h.

tan 30° = h/(8√3).

1/√3 = h/(8√3) ⇒ h = 8 m.

Option A.

Question 13

1 Mark Level 1 — Foundation NCERT Type Depression

From the top of a 24 m building, the angle of depression of a car on level ground is 45°. The car is horizontally how far from the building?

12 m

24 m

24√3 m

48 m

A 24 m building and a car seen at 45° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Angle of elevation from the car is also 45°.

tan 45° = 24/d ⇒ 1 = 24/d.

d = 24 m.

Option B.

Question 14

1 Mark Level 1 — Foundation NCERT Type Equation Selection

A tower of height h is viewed from a point d metres from its foot at angle of elevation θ. Which equation is correct?

tan θ = d/h

tan θ = h/d

sin θ = d/h

cos θ = h/d

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

The tower height is opposite to θ and the horizontal distance d is adjacent.

Therefore tan θ = opposite/adjacent = h/d.

Option B.

Question 15

1 Mark Level 1 — Foundation NCERT Type Error Analysis

A student writes tan 30° = 40/h for a tower of height h viewed from 40 m away. What is the error?

30° cannot be used

The ratio is inverted; it should be h/40

The height should be the hypotenuse

There is no error

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

For the observation angle, the opposite side is h and adjacent side is 40.

Thus tan 30° = h/40, not 40/h.

Option B.

Question 16

1 Mark Level 3 — Advanced Competency Based Moving Observer

A person is initially 36 m from a tower. The angle of elevation is 30°. After walking straight toward the tower, the angle becomes 60°. How far did the person walk?

12 m

18 m

24 m

30 m

Observer moves from 36 m away until the angle changes from 30° to 60°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: C

Let tower height be h.

Initially: tan 30° = h/36 ⇒ h = 36/√3 = 12√3 m.

Finally: tan 60° = h/x ⇒ √3 = 12√3/x ⇒ x = 12 m.

Distance walked = 36 − 12 = 24 m.

Option C.

Question 17

1 Mark Level 3 — Advanced Exemplar Type Opposite Sides

A tower of height 20√3 m stands between two observers. The angles of elevation at the two observers are 60° and 30°. The distance between the observers is:

40 m

60 m

80 m

100 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: C

Distance on 60° side: tan 60° = 20√3/x ⇒ x = 20 m.

Distance on 30° side: tan 30° = 20√3/y ⇒ y = 60 m.

Observers are on opposite sides, so total separation = 20 + 60 = 80 m.

Option C.

Question 18

1 Mark Level 3 — Advanced CBSE/PYQ Pattern Flagstaff

From a point 20 m from a building, the angle of elevation to the roof is 45° and to the top of a flagstaff on the roof is 60°. The flagstaff height is:

20√3 m

20(√3 − 1) m

20(√3 + 1) m

20 m

Building and flagstaff seen at 45° and 60°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Building height = 20 tan 45° = 20 m.

Total height to flagstaff top = 20 tan 60° = 20√3 m.

Flagstaff height = 20√3 − 20 = 20(√3 − 1) m.

Option B.

Question 19

1 Mark Level 1 — Foundation NCERT Exemplar Type Depression

From the top of a 15 m building, the angle of depression of a point on the ground is 30°. The horizontal distance to the point is:

5√3 m

15 m

15√3 m

30√3 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: C

Corresponding angle of elevation at ground = 30°.

tan 30° = 15/d ⇒ 1/√3 = 15/d.

d = 15√3 m.

Option C.

Question 20

1 Mark Level 3 — Advanced CBSE/PYQ Pattern Two Buildings

From the top of a 20 m building, the foot of another building is seen at 45° depression and its top at 45° elevation. The second building’s height is:

20 m

30 m

40 m

20√2 m

Top of one building sees foot and top of another at 45°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: C

From the depression triangle: tan 45° = 20/d ⇒ d = 20 m.

Height of second building above the observer’s level = d tan 45° = 20 m.

Second building height = 20 m below eye level + 20 m above eye level = 40 m.

Option C.

Question 21

1 Mark Level 1 — Foundation Competency Based Assertion–Reason

Assertion (A): The angle of depression of a boat from a lighthouse equals the angle of elevation of the lighthouse top from the boat.
Reason (R): The horizontal through the lighthouse top is parallel to the horizontal sea level.

Both A and R are true, and R correctly explains A

Both A and R are true, but R does not explain A

A is true but R is false

A is false but R is true

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: A

Both horizontals are parallel.

The line of sight is a transversal, giving equal alternate interior acute angles.

So R is precisely the geometric reason for A.

Option A.

Question 22

1 Mark Level 1 — Foundation Competency Based Assertion–Reason

Assertion (A): For a fixed tower height, the angle of elevation decreases as an observer walks away on level ground.
Reason (R): tan θ = h/d, so increasing d decreases tan θ.

Both A and R are true, and R correctly explains A

Both A and R are true, but R does not explain A

A is true but R is false

A is false but R is true

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: A

For fixed h, tan θ varies inversely with horizontal distance d.

For acute angles, a smaller tan value corresponds to a smaller θ.

Both statements are true and R explains A: option A.

Question 23

1 Mark Level 3 — Advanced HOTS Diagram Interpretation

Two points P and Q lie on the same side of a tower, with P nearer the tower than Q. If the angles of elevation are α at P and β at Q, which must be true?

α < β

α = β

α > β

α + β = 90°

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: C

For fixed tower height, tan(angle) = height/horizontal distance.

P has the smaller horizontal distance, so tan α > tan β.

For acute angles this implies α > β.

Option C.

Question 24

1 Mark Level 1 — Foundation Exemplar Type Line of Sight

The line of sight from an observer to the top of a pole is 10 m and makes a 30° angle with the horizontal. The vertical height above the observer’s eye level is:

5 m

5√3 m

10/√3 m

10 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: A

Here the line of sight is the hypotenuse.

sin 30° = h/10 ⇒ 1/2 = h/10.

h = 5 m.

Option A.

Question 25

1 Mark Level 1 — Foundation Exemplar Type Line of Sight

A cable from the top of a mast to a ground point has length 14√2 m and makes a 45° angle with the ground. The horizontal distance from the mast to the ground point is:

7 m

14 m

14√2 m

28 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Cable is the hypotenuse and horizontal distance d is adjacent to 45°.

cos 45° = d/(14√2).

1/√2 = d/(14√2) ⇒ d = 14 m.

Option B.

Question 26

1 Mark Level 1 — Foundation NCERT Type Exact Values

A calculation gives the height of a tower as 10√3 m. Unless a decimal approximation is explicitly requested, the best final form is:

17.32 m only

10√3 m

30 m

10/√3 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Standard-angle trigonometry should normally be kept in exact surd form.

10√3 m is exact; 17.32 m is approximate.

Option B.

Question 27

1 Mark Level 4 — HOTS HOTS Two Observation Points

From two points on the same side of a tower, 20 m apart, the angles of elevation are 45° at the nearer point and 30° at the farther point. The tower height is:

10(√3 − 1) m

10(√3 + 1) m

20√3 m

20 m

Same-side observation points 20 m apart with angles 45° and 30°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Let the nearer distance be x and height be h.

At 45°: tan 45° = h/x ⇒ h = x.

At 30°: tan 30° = h/(x + 20) ⇒ 1/√3 = x/(x + 20).

x + 20 = x√3 ⇒ x(√3 − 1) = 20.

x = 20/(√3 − 1) = 10(√3 + 1). Therefore h has the same value.

Option B.

Question 28

1 Mark Level 1 — Foundation CBSE/PYQ Pattern Lighthouse

A lighthouse is 30√3 m high. A ship is observed at an angle of depression 60°. The ship’s horizontal distance from the lighthouse foot is:

10 m

30 m

30√3 m

90 m

Ship viewed from a 30√3 m lighthouse at 60° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Angle of elevation from the ship = 60°.

tan 60° = 30√3/d.

√3 = 30√3/d ⇒ d = 30 m.

Option B.

Question 29

1 Mark Level 3 — Advanced Competency Based Observer Height

An observer’s eye is 1.6 m above the ground. A tower is 15√3 m away and its top is seen at 30°. The tower’s total height is:

15 m

16.6 m

15√3 + 1.6 m

18.2 m

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Height above eye level = h.

tan 30° = h/(15√3).

1/√3 = h/(15√3) ⇒ h = 15 m.

Total height = 15 + 1.6 = 16.6 m.

Option B.

Question 30

1 Mark Level 4 — HOTS Olympiad Flagstaff

A 10 m high building has a vertical antenna on its roof. From a ground point 10 m from the building, the angle to the roof is 45° and to the antenna top is 60°. Which expression gives the antenna height?

10√3

10(√3 − 1)

10(√3 + 1)

10/√3

A 10 m building with antenna viewed from 10 m away.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Correct option: B

Roof height check: 10 tan 45° = 10 m, consistent with the given building.

Total height to antenna top = 10 tan 60° = 10√3 m.

Antenna height = 10√3 − 10 = 10(√3 − 1) m.

Option B.

Section B — 15 Two-Mark Questions

Question 31

2 Marks Level 1 — Foundation NCERT Type Tower Height

From a point 25 m from the foot of a vertical tower, the angle of elevation of its top is 45°. Find the tower’s height.

Tower viewed from 25 m away at 45°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the height be h.

tan 45° = h/25.

1 = h/25 ⇒ h = 25 m.

Height of the tower = 25 m.

Question 32

2 Marks Level 1 — Foundation NCERT Type Horizontal Distance

A 12√3 m high pole is viewed at an angle of elevation 60°. Find the horizontal distance of the observer from its foot.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the distance be d.

tan 60° = 12√3/d.

√3 = 12√3/d ⇒ d = 12 m.

Horizontal distance = 12 m.

Question 33

2 Marks Level 1 — Foundation NCERT Type Elevation

A tree is 18 m from an observer. Its top is seen at 30°. Find the height of the tree above the observer’s eye level.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the required height be h.

tan 30° = h/18 ⇒ 1/√3 = h/18.

h = 18/√3 = 6√3 m.

Height above eye level = 6√3 m.

Question 34

2 Marks Level 1 — Foundation NCERT Type Depression

From the top of an 18 m building, the angle of depression of a bicycle on level ground is 45°. Find its distance from the foot of the building.

Bicycle viewed from an 18 m building at 45° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Angle of elevation from the bicycle = 45°.

Let horizontal distance = d. Then tan 45° = 18/d.

d = 18 m.

Distance from the building = 18 m.

Question 35

2 Marks Level 2 — Standard Competency Based Observer Height

A student’s eye level is 1.4 m above ground. From 12 m away, the top of a pole is seen at 45°. Find the total height of the pole.

Student eye level 1.4 m above ground viewing a pole.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Height of pole above eye level = h.

tan 45° = h/12 ⇒ h = 12 m.

Total height = 12 + 1.4 = 13.4 m.

Total pole height = 13.4 m.

Question 36

2 Marks Level 2 — Standard Exemplar Type Shadow

A vertical post casts a shadow 6 m long when the sun’s elevation is 60°. Find the post’s height.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let height = h.

tan 60° = h/6 ⇒ √3 = h/6.

h = 6√3 m.

Height of post = 6√3 m.

Question 37

2 Marks Level 2 — Standard Competency Based Line of Sight

The line of sight to the top of a wall is 16 m and makes a 30° angle with the horizontal. Find the vertical height above the observer’s eye level.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Here line of sight is the hypotenuse.

sin 30° = h/16 ⇒ 1/2 = h/16.

h = 8 m.

Vertical height = 8 m.

Question 38

2 Marks Level 2 — Standard NCERT Type Ratio Selection

In a tower problem, the tower height is 9 m and the horizontal distance is 9√3 m. Find the angle of elevation.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

tan θ = 9/(9√3) = 1/√3.

Since tan 30° = 1/√3, θ = 30°.

Angle of elevation = 30°.

Question 39

2 Marks Level 2 — Standard CBSE/PYQ Pattern Depression

A 10√3 m high platform overlooks a point on level ground at an angle of depression 30°. Find the horizontal distance.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Corresponding angle of elevation at the ground point = 30°.

tan 30° = 10√3/d ⇒ 1/√3 = 10√3/d.

d = 30 m.

Horizontal distance = 30 m.

Question 40

2 Marks Level 2 — Standard Exemplar Type Exact Values

Show that 24/√3 can be written without a surd in the denominator, and hence state its simplified exact value.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

24/√3 × √3/√3 = 24√3/3.

24√3/3 = 8√3.

Therefore, 24/√3 = 8√3.

Question 41

2 Marks Level 2 — Standard Competency Based Diagram Interpretation

In a right-triangle model of a tower, the horizontal distance is labelled PQ, the tower height QR, and the line of sight PR. The angle of elevation is at P. Identify the perpendicular, base and hypotenuse relative to the angle at P.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Relative to angle P: side opposite P is QR, so perpendicular = QR.

Side adjacent to P but not the hypotenuse is PQ, so base = PQ.

The side opposite the right angle is PR, so hypotenuse = PR.

P = QR, B = PQ, H = PR.

Question 42

2 Marks Level 2 — Standard NCERT Type Shadow

A pole is 10 m high. At what angle of elevation of the sun will its shadow also be 10 m long?

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

tan θ = height/shadow = 10/10 = 1.

Since tan 45° = 1, θ = 45°.

Sun’s angle of elevation = 45°.

Question 43

2 Marks Level 2 — Standard Competency Based Error Analysis

A learner says: “A 30° angle of depression should be used as 60° at the ground because the two angles are complementary.” Explain why this is incorrect.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

The horizontal through the observer and the ground horizontal are parallel.

The line of sight forms equal alternate interior angles, not complementary angles.

So the corresponding ground-level angle of elevation is also 30°.

Depression 30° corresponds to elevation 30°, not 60°.

Question 44

2 Marks Level 2 — Standard CBSE/PYQ Pattern Observer Height

A 1.5 m tall observer sees the top of a wall at 60° from a distance of 4√3 m. Find the wall height.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Height above eye level = h.

tan 60° = h/(4√3) ⇒ √3 = h/(4√3) ⇒ h = 12 m.

Total wall height = 12 + 1.5 = 13.5 m.

Wall height = 13.5 m.

Question 45

2 Marks Level 2 — Standard Exemplar Type Line of Sight

A supporting wire 20 m long is tied from the top of a pole to a ground point and makes 60° with the ground. Find the pole height.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Wire is the hypotenuse.

sin 60° = h/20 ⇒ √3/2 = h/20.

h = 10√3 m.

Pole height = 10√3 m.

Section C — 15 Three-Mark Questions

Question 46

3 Marks Level 2 — Standard NCERT Type Observer Height

A person whose eye level is 1.6 m above the ground stands 10√3 m from a tower. The angle of elevation of the top is 30°. Find the total height of the tower.

Observer 1.6 m high viewing a tower from 10√3 m at 30°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the height of the tower above eye level be h.

tan 30° = h/(10√3).

1/√3 = h/(10√3) ⇒ h = 10 m.

Total tower height = 10 + 1.6 = 11.6 m.

Height of tower = 11.6 m.

Question 47

3 Marks Level 2 — Standard NCERT Type Building Height

From a point 30 m from a building, the angle of elevation of its top is 60°. Find the building’s height and the length of the line of sight.

Building viewed from 30 m away at 60°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let building height = h.

tan 60° = h/30 ⇒ √3 = h/30 ⇒ h = 30√3 m.

Let line of sight = ℓ. Using cos 60° = 30/ℓ:

1/2 = 30/ℓ ⇒ ℓ = 60 m.

Building height = 30√3 m; line of sight = 60 m.

Question 48

3 Marks Level 2 — Standard CBSE/PYQ Pattern Lighthouse

From the top of a 36 m lighthouse, a boat is seen at an angle of depression 45°. Find (i) its horizontal distance from the lighthouse and (ii) its line-of-sight distance from the top.

Boat viewed from a 36 m lighthouse at 45° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Angle of elevation from the boat = 45°.

Let horizontal distance = d. tan 45° = 36/d ⇒ d = 36 m.

Let line of sight = ℓ. sin 45° = 36/ℓ.

1/√2 = 36/ℓ ⇒ ℓ = 36√2 m.

(i) 36 m (ii) 36√2 m.

Question 49

3 Marks Level 2 — Standard Exemplar Type Flagstaff

A building is 12 m high. From a point on the ground 12 m from its foot, the top of a flagstaff on the building is seen at 60°. Find the height of the flagstaff.

Flagstaff on a 12 m building viewed from 12 m away at 60°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Horizontal distance = 12 m.

Total height up to flagstaff top = H.

tan 60° = H/12 ⇒ H = 12√3 m.

Flagstaff height = total height − building height = 12√3 − 12.

Flagstaff height = 12(√3 − 1) m.

Question 50

3 Marks Level 2 — Standard Competency Based Moving Observer

A student is 60 m from a tower and sees its top at 30°. He walks 40 m straight toward the tower. Find the new angle of elevation.

Observer moves 40 m toward a tower from an initial distance of 60 m.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Initial tower height h = 60 tan 30° = 60/√3 = 20√3 m.

New horizontal distance = 60 − 40 = 20 m.

tan θ = h/20 = 20√3/20 = √3.

Since tan 60° = √3, θ = 60°.

New angle of elevation = 60°.

Question 51

3 Marks Level 2 — Standard CBSE/PYQ Pattern Two Observation Points

Two points A and B lie on the same straight line with the foot of a tower. B is 10 m nearer the tower than A. The angles of elevation from A and B are 30° and 45° respectively. Find the tower height.

Same-side points 10 m apart with angles 30° and 45°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the distance from B to the tower be x m and tower height be h.

From B: tan 45° = h/x ⇒ h = x. …(1)

From A: distance = x + 10. Hence tan 30° = h/(x + 10).

1/√3 = x/(x + 10) ⇒ x + 10 = x√3.

x(√3 − 1) = 10 ⇒ x = 10/(√3 − 1) = 5(√3 + 1).

Tower height h = 5(√3 + 1) m.

Question 52

3 Marks Level 3 — Advanced Exemplar Type Depression

A balcony is 15√3 m above level ground. Two points P and Q lie on the same side of the building, and their angles of depression are 60° and 30°. Find the distance PQ.

Two ground points on the same side viewed at 60° and 30° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal distances from the building foot be x for the 60° point and y for the 30° point.

tan 60° = 15√3/x ⇒ √3 = 15√3/x ⇒ x = 15 m.

tan 30° = 15√3/y ⇒ 1/√3 = 15√3/y ⇒ y = 45 m.

Both points are on the same side, so PQ = y − x = 45 − 15 = 30 m.

PQ = 30 m.

Question 53

3 Marks Level 3 — Advanced Competency Based Shadow

A 9 m pole casts a shadow 3√3 m long. Later, its shadow becomes 9√3 m long. Find the sun’s angle of elevation in each situation.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

First situation: tan θ₁ = 9/(3√3) = √3 ⇒ θ₁ = 60°.

Second situation: tan θ₂ = 9/(9√3) = 1/√3 ⇒ θ₂ = 30°.

The longer shadow corresponds to the smaller solar elevation angle.

Angles are 60° first and 30° later.

Question 54

3 Marks Level 3 — Advanced CBSE/PYQ Pattern Top and Bottom

From the top of a 24 m building, the foot of another vertical pole is seen at 45° depression and its top at 30° depression. Find the pole height.

Top of a 24 m building sees the foot at 45° depression and pole top at 30° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal distance between building and pole be d.

Using the foot: tan 45° = 24/d ⇒ d = 24 m.

Let the vertical drop from building top to pole top be x.

tan 30° = x/24 ⇒ x = 24/√3 = 8√3 m.

Pole height = 24 − x = 24 − 8√3 m.

Pole height = 24 − 8√3 m.

Question 55

3 Marks Level 3 — Advanced Competency Based Drone

A drone hovers vertically above a point on level ground. From an observer 50 m away from that point, the drone is seen at 45°. If the observer’s eye is 1.5 m above ground, find the drone’s altitude above ground.

Drone directly above a ground point, viewed from 50 m away at 45°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Height of drone above eye level = h.

tan 45° = h/50 ⇒ h = 50 m.

Altitude above ground = 50 + 1.5 = 51.5 m.

Drone altitude = 51.5 m.

Question 56

3 Marks Level 3 — Advanced Exemplar Type Two Buildings

From the top of a 20 m building, the top of a 20√3 m building is seen at an angle of elevation 30°. Find the horizontal distance between the buildings.

Top of the shorter building views the taller building top at 30° elevation.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Difference in heights = 20√3 − 20 = 20(√3 − 1) m.

Let horizontal distance = d.

tan 30° = 20(√3 − 1)/d.

1/√3 = 20(√3 − 1)/d ⇒ d = 20√3(√3 − 1) = 20(3 − √3) m.

Horizontal distance = 20(3 − √3) m.

Question 57

3 Marks Level 3 — Advanced Competency Based Bridge

From a point on a bridge 18 m above river level, the angle of depression of a buoy is 60°. Find the horizontal distance to the buoy and the line-of-sight distance.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Angle of elevation from the buoy = 60°.

Let horizontal distance = d. tan 60° = 18/d ⇒ d = 18/√3 = 6√3 m.

Let line of sight = ℓ. sin 60° = 18/ℓ ⇒ √3/2 = 18/ℓ.

ℓ = 36/√3 = 12√3 m.

Horizontal distance = 6√3 m; line of sight = 12√3 m.

Question 58

3 Marks Level 3 — Advanced HOTS Equation Selection

A tower is observed from a point P at 60°. A second point Q is 16 m farther from the tower along the same line and the angle at Q is 30°. Without solving completely, form the two equations needed to determine the tower height h and nearer distance x.

Two observation points separated by 16 m, with angles 60° and 30°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let nearer distance from P be x m. Then farther distance from Q is x + 16 m.

From P: tan 60° = h/x ⇒ √3 = h/x. …(1)

From Q: tan 30° = h/(x + 16) ⇒ 1/√3 = h/(x + 16). …(2)

Required equations: h = x√3 and h = (x + 16)/√3.

Question 59

3 Marks Level 3 — Advanced CBSE/PYQ Pattern Flagstaff

From a point on level ground, the angle of elevation to the roof of a building is 30°, while that to the top of a 6 m flagstaff on the roof is 45°. Find the building height.

Building with a 6 m flagstaff viewed at 30° to roof and 45° to flag top.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal distance be d and building height be h.

To roof: tan 30° = h/d ⇒ d = h√3. …(1)

To flagstaff top: tan 45° = (h + 6)/d ⇒ d = h + 6. …(2)

Equating: h√3 = h + 6 ⇒ h(√3 − 1) = 6.

h = 6/(√3 − 1) = 3(√3 + 1) m.

Building height = 3(√3 + 1) m.

Question 60

3 Marks Level 3 — Advanced HOTS Opposite Sides

A tower stands between two points P and Q that are 70 m apart. The angles of elevation of the top are 45° at P and 30° at Q. Find the distance of the tower from P.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let distance from P to tower = x m, so distance from Q = 70 − x m. Let height = h.

From P: tan 45° = h/x ⇒ h = x.

From Q: tan 30° = h/(70 − x) ⇒ 1/√3 = x/(70 − x).

70 − x = x√3 ⇒ 70 = x(√3 + 1).

x = 70/(√3 + 1) = 35(√3 − 1) m.

Distance from P = 35(√3 − 1) m.

Section D — 20 Four-Mark Questions

Question 61

4 Marks Level 3 — Advanced CBSE/PYQ Pattern Two Observation Points

Two points A and B are on the same side of a tower and 40 m apart. The angle of elevation of the tower top is 60° from the nearer point A and 30° from the farther point B. Find the tower height and the distance of A from the tower.

Same-side points 40 m apart with angles 60° and 30°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the distance from A to the tower be x m and tower height be h.

From A: tan 60° = h/x ⇒ √3 = h/x ⇒ h = x√3. …(1)

Since B is 40 m farther, its distance from the tower is x + 40.

From B: tan 30° = h/(x + 40) ⇒ 1/√3 = h/(x + 40). …(2)

Using h = x√3 in (2): 1/√3 = x√3/(x + 40).

x + 40 = 3x ⇒ 2x = 40 ⇒ x = 20 m.

h = 20√3 m.

Distance of A = 20 m; tower height = 20√3 m.

Question 62

4 Marks Level 3 — Advanced Competency Based Moving Observer

A cyclist sees the top of a tower at 60°. After riding 30 m directly away from the tower on level ground, the angle becomes 30°. Find the original distance from the tower and the tower height.

Cyclist moves 30 m away; angle changes from 60° to 30°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let original distance = x m and tower height = h.

Initially: tan 60° = h/x ⇒ h = x√3. …(1)

After moving away, distance = x + 30.

tan 30° = h/(x + 30) ⇒ h = (x + 30)/√3. …(2)

Equate (1) and (2): x√3 = (x + 30)/√3.

3x = x + 30 ⇒ 2x = 30 ⇒ x = 15 m.

h = 15√3 m.

Original distance = 15 m; tower height = 15√3 m.

Question 63

4 Marks Level 3 — Advanced CBSE/PYQ Pattern Two Buildings

From the top of a 30 m building, the foot of a second building is seen at an angle of depression 45°, while its top is seen at an angle of elevation 30°. Find the horizontal distance between the buildings and the height of the second building.

A 30 m building views the foot at 45° depression and top at 30° elevation.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal distance = d.

From the depression triangle: tan 45° = 30/d ⇒ d = 30 m.

Let the second building extend x m above the top level of the first building.

From the elevation triangle: tan 30° = x/30.

1/√3 = x/30 ⇒ x = 10√3 m.

Second building height = 30 + 10√3 m.

Distance = 30 m; second building height = 30 + 10√3 m.

Question 64

4 Marks Level 3 — Advanced Exemplar Type Flagstaff

From a point 24 m from a building, the roof is seen at 45° and the top of a vertical flagstaff on the roof at 60°. Find the building height, the total height up to the flagstaff top, and the flagstaff height.

Building roof at 45° and flagstaff top at 60° from 24 m away.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let building height = h and total height = H.

To the roof: tan 45° = h/24 ⇒ h = 24 m.

To the flagstaff top: tan 60° = H/24 ⇒ H = 24√3 m.

Flagstaff height = H − h = 24√3 − 24.

Building = 24 m; total = 24√3 m; flagstaff = 24(√3 − 1) m.

Question 65

4 Marks Level 3 — Advanced CBSE/PYQ Pattern Lighthouse

A lighthouse is 30√3 m high. Two ships are on the same side of it and are observed at angles of depression 60° and 30°. Find their distances from the lighthouse and the distance between the ships.

Two ships on the same side seen at 60° and 30° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the nearer ship be at distance x and farther ship at distance y.

For 60°: tan 60° = 30√3/x ⇒ √3 = 30√3/x ⇒ x = 30 m.

For 30°: tan 30° = 30√3/y ⇒ 1/√3 = 30√3/y ⇒ y = 90 m.

Same side ⇒ separation = y − x = 90 − 30 = 60 m.

Distances are 30 m and 90 m; ships are 60 m apart.

Question 66

4 Marks Level 3 — Advanced Exemplar Type Opposite Sides

Two observers are 80 m apart with a tower between them. The angles of elevation of the top are 30° and 60°. Find the tower height and its distance from each observer.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let tower height = h. Let distances on the 30° and 60° sides be x and y respectively.

At 30°: tan 30° = h/x ⇒ x = h√3.

At 60°: tan 60° = h/y ⇒ y = h/√3.

Since tower lies between them: x + y = 80.

h√3 + h/√3 = 80 ⇒ h(3 + 1)/√3 = 80.

4h/√3 = 80 ⇒ h = 20√3 m.

Then x = 20√3·√3 = 60 m and y = 20√3/√3 = 20 m.

Tower height = 20√3 m; distances = 60 m and 20 m.

Question 67

4 Marks Level 3 — Advanced Competency Based Observer Height

A surveyor whose instrument is 1.5 m above ground observes the top of a tower from two points on the same line with its foot. At 15 m from the foot the angle is 45°; at a farther point the angle is 30°. Find the tower’s total height and the distance between the two observation points.

Survey instrument 1.5 m high used from two ground positions.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the portion of tower above instrument level be h.

At the nearer point: tan 45° = h/15 ⇒ h = 15 m.

Total tower height = 15 + 1.5 = 16.5 m.

Let farther horizontal distance from tower = d.

tan 30° = h/d ⇒ 1/√3 = 15/d ⇒ d = 15√3 m.

Distance between observation points = 15√3 − 15 = 15(√3 − 1) m.

Tower height = 16.5 m; point separation = 15(√3 − 1) m.

Question 68

4 Marks Level 3 — Advanced CBSE/PYQ Pattern Two Buildings

From the top of a 24 m building, the foot of another building is seen at 30° depression and its top at 30° elevation. Find the horizontal distance between the buildings and the height of the second building.

Equal 30° elevation and depression from the top of a 24 m building.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal distance = d.

Depression 30° gives: tan 30° = 24/d.

1/√3 = 24/d ⇒ d = 24√3 m.

Let height of second building above the observer’s level be x.

Elevation 30° gives: tan 30° = x/d ⇒ x = d/√3 = 24 m.

Second building height = 24 + 24 = 48 m.

Distance = 24√3 m; second building height = 48 m.

Question 69

4 Marks Level 3 — Advanced Competency Based Shadow

A 12√3 m vertical mast casts a 12 m shadow at one time and a 36 m shadow later. Find the sun’s elevation angle in each case and determine by how much the shadow length has increased.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

First: tan θ₁ = 12√3/12 = √3 ⇒ θ₁ = 60°.

Later: tan θ₂ = 12√3/36 = 1/√3 ⇒ θ₂ = 30°.

Increase in shadow length = 36 − 12 = 24 m.

The fall in solar elevation from 60° to 30° explains the longer shadow.

Angles: 60° and 30°; shadow increases by 24 m.

Question 70

4 Marks Level 4 — HOTS HOTS Cliff and Boat

From the top of a vertical cliff, a boat is first seen at an angle of depression 30°. After the boat moves 40 m straight toward the cliff, the angle becomes 45°. Find the height of the cliff.

Boat moves toward a cliff; depression changes from 30° to 45°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let cliff height = h and the boat’s nearer distance after moving be x.

At the nearer position: tan 45° = h/x ⇒ x = h.

At the initial position, horizontal distance = x + 40 = h + 40.

tan 30° = h/(h + 40).

1/√3 = h/(h + 40) ⇒ h + 40 = h√3.

h(√3 − 1) = 40.

h = 40/(√3 − 1) = 20(√3 + 1) m.

Cliff height = 20(√3 + 1) m.

Question 71

4 Marks Level 4 — HOTS HOTS Two Observation Points

A tower is viewed from two points on the same side. At the nearer point the angle of elevation is 45°; at the farther point it is 30°. If the tower is 10(√3 + 1) m high, find the distance between the two points.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let height h = 10(√3 + 1) m.

At nearer point (45°): distance x = h/tan45° = h = 10(√3 + 1).

At farther point (30°): distance y = h/tan30° = h√3.

y = 10(√3 + 1)√3 = 10(3 + √3).

Separation = y − x = 10(3 + √3) − 10(√3 + 1) = 20 m.

Distance between observation points = 20 m.

Question 72

4 Marks Level 4 — HOTS Competency Based Stadium Light

A stadium floodlight is mounted on top of a 20 m stand. From a ground point, the angle of elevation of the top of the stand is 45° and that of the lamp is 60°. Find the horizontal distance to the stand and the height of the lamp above the stand.

Floodlight above a 20 m stand, viewed at 45° and 60°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal distance = d.

To top of stand: tan 45° = 20/d ⇒ d = 20 m.

Let total height to lamp = H.

tan 60° = H/20 ⇒ H = 20√3 m.

Lamp height above stand = H − 20 = 20(√3 − 1) m.

Horizontal distance = 20 m; lamp is 20(√3 − 1) m above the stand.

Question 73

4 Marks Level 4 — HOTS HOTS Two Buildings

A 25 m building and a taller building stand on level ground. From the top of the shorter building, the foot of the taller one is seen at 45° depression and its top at 60° elevation. Find the taller building’s height.

Shorter building sees foot at 45° depression and top at 60° elevation.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal separation = d.

From the foot of the taller building: tan 45° = 25/d ⇒ d = 25 m.

Let height difference above the 25 m level be x.

tan 60° = x/25 ⇒ x = 25√3 m.

Total taller-building height = 25 + 25√3 = 25(1 + √3) m.

Height of taller building = 25(1 + √3) m.

Question 74

4 Marks Level 4 — HOTS Olympiad Opposite Sides

A tower stands between points A and B. The angles of elevation at A and B are 60° and 45° respectively. If AB = 30(√3 + 1) m, find the tower height.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let distances from tower foot to A and B be x and y, and height be h.

At A: tan 60° = h/x ⇒ x = h/√3.

At B: tan 45° = h/y ⇒ y = h.

Since the tower lies between A and B: x + y = 30(√3 + 1).

h/√3 + h = 30(√3 + 1).

h(1 + √3)/√3 = 30(√3 + 1).

Cancel (√3 + 1): h/√3 = 30 ⇒ h = 30√3 m.

Tower height = 30√3 m.

Question 75

4 Marks Level 4 — HOTS Olympiad Moving Observer

An observer sees the top of a tower at 30°. After moving 20 m toward it, the angle becomes 45°. He moves further toward the tower until the angle becomes 60°. Find the additional distance travelled in the second movement.

Three observer positions corresponding to 30°, 45° and 60°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let tower height = h.

At 45° position, horizontal distance = h.

At 30° initial position, distance = h√3.

Given first movement: h√3 − h = 20 ⇒ h(√3 − 1) = 20.

So h = 20/(√3 − 1) = 10(√3 + 1).

At 60° final position, distance = h/√3.

Second movement = h − h/√3 = h(1 − 1/√3).

Substitute h = 10(√3 + 1): second movement = 10(√3 + 1)(√3 − 1)/√3 = 20/√3 = 20√3/3 m.

Additional distance = 20√3/3 m.

Question 76

4 Marks Level 4 — HOTS Olympiad Lighthouse

Two ships lie on opposite sides of a lighthouse of height 40 m. Their angles of depression are 30° and 45°. Find the distance between the ships.

Ships on opposite sides of a lighthouse, seen at 30° and 45° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Horizontal distance to the 30° ship: d₁ = 40/tan30° = 40√3 m.

Horizontal distance to the 45° ship: d₂ = 40/tan45° = 40 m.

The ships are on opposite sides, so their separation is d₁ + d₂.

Distance = 40√3 + 40 = 40(√3 + 1) m.

Distance between ships = 40(√3 + 1) m.

Question 77

4 Marks Level 5 — Olympiad Challenge Olympiad Two Towers

Two vertical towers have heights 20 m and 20√3 m. From the top of the shorter tower, the angle of elevation of the taller tower’s top is 30°. Find the horizontal distance between the towers. If φ is the angle of depression of the taller tower’s foot from the top of the shorter tower, find tan φ.

Two towers of unequal heights with an elevation to the taller top.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Height difference = 20√3 − 20 = 20(√3 − 1) m.

Let horizontal distance = d.

tan 30° = 20(√3 − 1)/d.

d = 20√3(√3 − 1) = 20(3 − √3) m.

For the depression angle φ to the taller tower’s foot: tan φ = 20/d.

tan φ = 20/[20(3 − √3)] = 1/(3 − √3).

Rationalising: tan φ = (3 + √3)/6.

Distance = 20(3 − √3) m; tan φ = (3 + √3)/6.

Question 78

4 Marks Level 5 — Olympiad Challenge Olympiad Observer Height

A camera is mounted 2 m above the ground. It views the top of a vertical monument at 45°. After moving 10 m farther away on the same straight line, the angle becomes 30°. Find the monument’s total height.

A 2 m camera observes a monument from two positions 10 m apart.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let monument height above camera level = h and nearer horizontal distance = x.

At nearer point: tan 45° = h/x ⇒ h = x.

At farther point: distance = x + 10 and tan 30° = h/(x + 10).

1/√3 = x/(x + 10) ⇒ x + 10 = x√3.

x(√3 − 1) = 10 ⇒ x = 5(√3 + 1) m.

Thus height above camera level = 5(√3 + 1) m.

Total monument height = 5(√3 + 1) + 2 m.

Total height = 7 + 5√3 m.

Question 79

4 Marks Level 5 — Olympiad Challenge Olympiad Top and Bottom

From the top of a tower, the angles of depression of the top and foot of a 12 m pole standing on the same level ground are 30° and 45° respectively. Find the tower height.

Tower top views both top and foot of a 12 m pole.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let tower height = H and horizontal distance between tower and pole = d.

To pole foot: tan 45° = H/d ⇒ d = H.

Vertical drop from tower top to pole top = H − 12.

To pole top: tan 30° = (H − 12)/d.

1/√3 = (H − 12)/H.

H = √3(H − 12) ⇒ H(√3 − 1) = 12√3.

H = 12√3/(√3 − 1) = 6√3(√3 + 1) = 18 + 6√3 m.

Tower height = 18 + 6√3 m.

Question 80

4 Marks Level 5 — Olympiad Challenge Olympiad Geometric Constraint

A student proposes the following configuration: a vertical tower is seen from point P at 30° and from point Q at 60°, where P and Q are on the same side of the tower. The student also claims that the midpoint M of PQ sees the top at 45°. Test whether all three statements can be true. In the process, express PQ in terms of tower height h.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let distances of P and Q from the tower foot be p and q respectively, with P farther away.

From P: tan 30° = h/p ⇒ p = h√3.

From Q: tan 60° = h/q ⇒ q = h/√3.

Therefore PQ = p − q = h√3 − h/√3 = 2h/√3.

Midpoint distance from tower = (p + q)/2 = [h√3 + h/√3]/2 = 2h/√3.

Thus tan ∠M = h/(2h/√3) = √3/2.

But tan 45° = 1, whereas √3/2 ≠ 1.

Hence the 30° and 60° observations imply PQ = 2h/√3, but the additional claim ∠M = 45° is impossible. The proposed configuration is inconsistent.

Section E — 10 Five-Mark Questions

Question 81

5 Marks Level 4 — HOTS CBSE/PYQ Pattern Two Observation Points

Two observation points A and B lie on the same side of a tower, with A nearer the tower. They are 50 m apart. The angles of elevation of the tower top are 60° at A and 30° at B. Find (i) the distance of A from the tower, (ii) the tower height, and (iii) the line-of-sight distance from A to the top.

Same-side observers 50 m apart; angles 60° and 30°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let distance from A to the tower be x m and tower height be h.

From A: tan 60° = h/x ⇒ h = x√3. …(1)

From B: distance = x + 50 and tan 30° = h/(x + 50).

1/√3 = h/(x + 50) ⇒ h = (x + 50)/√3. …(2)

Equating (1) and (2): x√3 = (x + 50)/√3 ⇒ 3x = x + 50.

2x = 50 ⇒ x = 25 m.

Then h = 25√3 m.

Let line of sight from A be ℓ. sin 60° = h/ℓ.

√3/2 = 25√3/ℓ ⇒ ℓ = 50 m.

(i) 25 m (ii) 25√3 m (iii) 50 m.

Question 82

5 Marks Level 4 — HOTS CBSE/PYQ Pattern Two Buildings

From the top of a 30 m building, the foot of a taller building is seen at 45° depression and its top at 60° elevation. Find (i) the horizontal separation, (ii) the height of the taller building, and (iii) the straight-line distance from the observer to the taller building’s top.

A 30 m building views another building’s foot at 45° depression and top at 60° elevation.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal separation = d.

Using the foot: tan 45° = 30/d ⇒ d = 30 m.

Let the portion of the taller building above the observer’s level be x.

tan 60° = x/30 ⇒ x = 30√3 m.

Total height of taller building = 30 + 30√3 = 30(1 + √3) m.

Let line-of-sight distance to the top be ℓ.

cos 60° = 30/ℓ ⇒ 1/2 = 30/ℓ ⇒ ℓ = 60 m.

(i) 30 m (ii) 30(1 + √3) m (iii) 60 m.

Question 83

5 Marks Level 4 — HOTS Exemplar Type Building + Flagstaff

A vertical flagstaff 20 m high stands on the roof of a building. From a point on level ground, the angles of elevation of the roof and the top of the flagstaff are 30° and 60° respectively. Find the building height and the horizontal distance of the observation point from the building.

A 20 m flagstaff on a building, with 30° to roof and 60° to flag top.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let building height = h and horizontal distance = d.

To the roof: tan 30° = h/d.

1/√3 = h/d ⇒ h = d/√3. …(1)

To the flagstaff top: total height = h + 20.

tan 60° = (h + 20)/d ⇒ √3 = (h + 20)/d.

h + 20 = d√3. …(2)

Substitute h = d/√3 into (2): d/√3 + 20 = d√3.

20 = d(√3 − 1/√3) = 2d/√3.

d = 10√3 m.

Then h = d/√3 = 10 m.

Building height = 10 m; horizontal distance = 10√3 m.

Question 84

5 Marks Level 4 — HOTS Competency Based Observer Height + Motion

A surveyor’s instrument is 1.5 m above level ground. From a nearer point the top of a tower is seen at 60°. The surveyor then walks 20 m directly away from the tower and the angle becomes 30°. Find (i) the nearer horizontal distance, (ii) the height of the tower above instrument level, and (iii) the total tower height.

Survey instrument 1.5 m high used from two positions 20 m apart.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let nearer horizontal distance = x m and height above instrument level = h.

Nearer position: tan 60° = h/x ⇒ h = x√3. …(1)

Farther position: distance = x + 20.

tan 30° = h/(x + 20) ⇒ h = (x + 20)/√3. …(2)

Equating: x√3 = (x + 20)/√3.

3x = x + 20 ⇒ 2x = 20 ⇒ x = 10 m.

h = 10√3 m.

Total tower height = 10√3 + 1.5 m.

(i) 10 m (ii) 10√3 m (iii) 10√3 + 1.5 m.

Question 85

5 Marks Level 4 — HOTS HOTS Lighthouse and Ships

Two ships are on the same side of a lighthouse. From the lighthouse top, their angles of depression are 45° and 30°. The ships are 50 m apart along the same straight line from the lighthouse. Find the lighthouse height and each ship’s distance from its foot.

Two same-side ships 50 m apart, with depression angles 45° and 30°.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let lighthouse height = h.

The 45° ship is nearer. Its horizontal distance d₁ satisfies tan 45° = h/d₁ ⇒ d₁ = h.

For the 30° ship: tan 30° = h/d₂ ⇒ d₂ = h√3.

Since the ships are on the same side and 50 m apart: d₂ − d₁ = 50.

h√3 − h = 50 ⇒ h(√3 − 1) = 50.

h = 50/(√3 − 1) = 25(√3 + 1) m.

Nearer distance d₁ = 25(√3 + 1) m.

Farther distance d₂ = h√3 = 25(3 + √3) m.

Lighthouse height = 25(√3 + 1) m; ship distances = 25(√3 + 1) m and 25(3 + √3) m.

Question 86

5 Marks Level 5 — Olympiad Challenge Olympiad Cliff and Boats

Two boats lie on opposite sides of the foot of a vertical cliff. From the cliff top, their angles of depression are 45° and 30°. The boats are 100 m apart. Find the cliff height and the horizontal distance of each boat from the cliff foot.

Boats on opposite sides of a cliff, seen at 45° and 30° depression.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let cliff height = h.

For the 45° boat: tan 45° = h/d₁ ⇒ d₁ = h.

For the 30° boat: tan 30° = h/d₂ ⇒ d₂ = h√3.

The boats lie on opposite sides, so d₁ + d₂ = 100.

h + h√3 = 100 ⇒ h(1 + √3) = 100.

h = 100/(1 + √3) = 50(√3 − 1) m.

d₁ = 50(√3 − 1) m.

d₂ = h√3 = 50(3 − √3) m.

Check: d₁ + d₂ = 50[(√3 − 1) + (3 − √3)] = 100 m.

Cliff height = 50(√3 − 1) m; distances = 50(√3 − 1) m and 50(3 − √3) m.

Question 87

5 Marks Level 5 — Olympiad Challenge Olympiad Opposite Observers

Two observers, each with eye level 1.5 m above the ground, stand 70 m apart on opposite sides of a tower. The angles of elevation of the tower top are 45° and 30°. Find the total height of the tower and the ground distance from the tower to each observer.

Two equal-eye-height observers on opposite sides of a tower.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the height of tower above eye level be h.

Let the distance on the 45° side be x and on the 30° side be y.

At 45°: tan 45° = h/x ⇒ x = h.

At 30°: tan 30° = h/y ⇒ y = h√3.

Opposite sides ⇒ x + y = 70.

h + h√3 = 70 ⇒ h = 70/(1 + √3) = 35(√3 − 1) m.

Thus x = 35(√3 − 1) m.

y = h√3 = 35(3 − √3) m.

Total tower height = h + 1.5 = 35(√3 − 1) + 1.5 m.

Tower height = 35(√3 − 1) + 1.5 m; distances = 35(√3 − 1) m and 35(3 − √3) m.

Question 88

5 Marks Level 5 — Olympiad Challenge Olympiad Three Observation Positions

An observer walks straight toward a tower. The angle of elevation changes successively from 30° to 45° to 60°. The total distance walked from the 30° position to the 60° position is 50 m. Find the tower height and the distances walked in the two stages.

Three observation positions: 30°, 45° and 60° while approaching a tower.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let tower height = h.

At 30°: distance d₁ = h/tan30° = h√3.

At 45°: distance d₂ = h.

At 60°: distance d₃ = h/√3.

Total movement d₁ − d₃ = 50.

h√3 − h/√3 = 50 ⇒ h(3 − 1)/√3 = 50.

2h/√3 = 50 ⇒ h = 25√3 m.

Then d₁ = 75 m, d₂ = 25√3 m, d₃ = 25 m.

First stage distance = d₁ − d₂ = 75 − 25√3 = 25(3 − √3) m.

Second stage distance = d₂ − d₃ = 25√3 − 25 = 25(√3 − 1) m.

Tower height = 25√3 m; stage distances = 25(3 − √3) m and 25(√3 − 1) m.

Question 89

5 Marks Level 5 — Olympiad Challenge Olympiad Two Towers

From the top of a 24 m tower, the foot of a second tower is seen at 45° depression and the top of the second tower at 30° elevation. Find (i) the horizontal distance between the towers, (ii) the second tower’s height, and (iii) tan θ, where θ is the angle of elevation of the second tower’s top from the foot of the first tower.

A 24 m tower sees the second tower’s foot at 45° depression and top at 30° elevation.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal distance = d.

Using the depression to the second tower’s foot: tan 45° = 24/d ⇒ d = 24 m.

Let height of second tower above the first tower’s top level be x.

tan 30° = x/24 ⇒ x = 24/√3 = 8√3 m.

Second tower height = 24 + 8√3 m.

From the foot of the first tower, tan θ = (second tower height)/(horizontal distance).

tan θ = (24 + 8√3)/24 = 1 + √3/3 = (3 + √3)/3.

(i) 24 m (ii) 24 + 8√3 m (iii) tan θ = (3 + √3)/3.

Question 90

5 Marks Level 5 — Olympiad Challenge Olympiad Midpoint Reasoning

A tower stands between observers A and B, who are 120 m apart. The angle of elevation of the tower top is 30° from A and 60° from B. Find (i) the tower height, (ii) the distances from A and B to the tower, and (iii) the angle of elevation from the midpoint M of AB.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let tower height = h. Let distances from A and B to the tower be x and y.

At A: tan 30° = h/x ⇒ x = h√3.

At B: tan 60° = h/y ⇒ y = h/√3.

Since the tower is between A and B: x + y = 120.

h√3 + h/√3 = 120 ⇒ 4h/√3 = 120 ⇒ h = 30√3 m.

Then x = h√3 = 90 m and y = h/√3 = 30 m.

Midpoint M is 60 m from A. Since the tower is 90 m from A, M is 30 m from the tower.

Thus tan ∠M = h/30 = 30√3/30 = √3.

So ∠M = 60°.

(i) 30√3 m (ii) 90 m and 30 m (iii) 60°.

Section F — 5 Case Studies (4 Sub-questions Each)

Case Study 1 (Main Entry 91)

4 Sub-questions Level 3 — Advanced Competency Based Case Study

Lighthouse Navigation

A lighthouse is 30√3 m high. Two ships S₁ and S₂ are on the same side of the lighthouse. From the lighthouse top, the angles of depression are 60° to S₁ and 30° to S₂.

Lighthouse with two same-side ships at depression angles 60° and 30°.

(a) What is the angle of elevation of the lighthouse top from S₁?

(b) Find the horizontal distance of S₁ from the lighthouse foot.

(d) S₁ sails directly away from the lighthouse until its angle of elevation becomes 45°. How far does it sail?

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Complete Case-Study Solution

Part (a)

The sea level and the horizontal through the lighthouse top are parallel.

Therefore angle of elevation from S₁ = angle of depression = 60°.

Answer: 60°.

Part (b)

Let distance = x.

tan 60° = 30√3/x.

√3 = 30√3/x ⇒ x = 30 m.

S₁ is 30 m from the lighthouse.

Part (c)

Let S₂ distance = y.

tan 30° = 30√3/y ⇒ 1/√3 = 30√3/y ⇒ y = 90 m.

Same side ⇒ S₁S₂ = 90 − 30 = 60 m.

S₂ distance = 90 m; ship separation = 60 m.

Part (d)

At 45°, new distance d satisfies tan 45° = 30√3/d.

Thus d = 30√3 m.

Initial distance of S₁ = 30 m.

Distance sailed = 30√3 − 30 = 30(√3 − 1) m.

S₁ sails 30(√3 − 1) m.

Case Study 2 (Main Entry 92)

4 Sub-questions Level 3 — Advanced CBSE/PYQ Pattern Case Study

School Building and Flagstaff

A school building is 15 m high. A vertical flagstaff stands on its roof. From a point P on level ground 15 m from the building, the roof is seen at 45° and the flagstaff top at 60°.

School building 15 m high with a roof flagstaff; P is 15 m away.

(a) Verify the 45° observation to the roof using a trigonometric ratio.

(b) Find the total height from ground to the top of the flagstaff.

(d) If a second observation point Q is chosen so that the flagstaff top is seen at 45°, find Q’s distance from the building and compare it with P.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Complete Case-Study Solution

Part (a)

tan 45° = building height/horizontal distance = 15/15 = 1.

Since tan 45° = 1, the observation is consistent.

Verified.

Part (b)

Let total height = H.

tan 60° = H/15 ⇒ √3 = H/15.

H = 15√3 m.

Total height = 15√3 m.

Part (c)

Flagstaff height = total height − building height.

= 15√3 − 15.

Flagstaff height = 15(√3 − 1) m.

Part (d)

At Q, tan 45° = total height/d ⇒ 1 = 15√3/d.

So d = 15√3 m.

P is 15 m away, so Q is farther by 15√3 − 15 = 15(√3 − 1) m.

Q is 15√3 m from the building, i.e. 15(√3 − 1) m farther than P.

Case Study 3 (Main Entry 93)

4 Sub-questions Level 4 — HOTS Competency Based Case Study

Drone Survey

A survey drone hovers vertically above a marked point O. A surveyor’s eye is 1.5 m above level ground. From point A, 20 m from O, the drone is seen at an angle of elevation 45°.

Drone above O viewed from a surveyor with eye height 1.5 m.

(a) What height difference between the drone and the surveyor’s eye is represented by the perpendicular of the right triangle?

(b) Find the drone’s altitude above ground.

(d) How far must the surveyor move directly away from O from point A to reach that 30° position?

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Complete Case-Study Solution

Part (a)

Let the height difference be h.

tan 45° = h/20 ⇒ h = 20 m.

Height above eye level = 20 m.

Part (b)

Altitude = height above eye level + eye height.

= 20 + 1.5 = 21.5 m.

Drone altitude = 21.5 m.

Part (c)

The vertical height above eye level remains 20 m.

tan 30° = 20/d ⇒ 1/√3 = 20/d.

d = 20√3 m.

Required distance = 20√3 m.

Part (d)

Initial distance = 20 m.

New distance = 20√3 m.

Movement away = 20√3 − 20 = 20(√3 − 1) m.

The surveyor moves 20(√3 − 1) m.

Case Study 4 (Main Entry 94)

4 Sub-questions Level 4 — HOTS Exemplar Type Case Study

Fire Lookout Tower

A fire lookout tower is 20√3 m high. Two smoke locations A and B lie on opposite sides of the tower on level ground. Their angles of elevation to the lookout top are 60° and 30° respectively.

A lookout tower between two opposite-side smoke locations with angles 60° and 30°.

(a) Which smoke location is nearer the tower, and why?

(b) Find the distance from A to the tower.

(d) Find AB and explain why the distances are added rather than subtracted.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Complete Case-Study Solution

Part (a)

For a fixed height, the larger angle corresponds to the smaller horizontal distance.

Since 60° > 30°, location A (60°) is nearer.

A is nearer.

Part (b)

tan 60° = 20√3/x.

√3 = 20√3/x ⇒ x = 20 m.

A is 20 m from the tower.

Part (c)

tan 30° = 20√3/y.

1/√3 = 20√3/y ⇒ y = 60 m.

B is 60 m from the tower.

Part (d)

A and B are on opposite sides of the tower.

Therefore the segment AB passes through the tower foot.

AB = 20 + 60 = 80 m.

AB = 80 m; opposite-side distances are added.

Case Study 5 (Main Entry 95)

4 Sub-questions Level 5 — Olympiad Challenge Competency Based Case Study

Construction Survey Between Two Buildings

A surveyor stands at the top of an 18 m building. The foot of a second building is seen at an angle of depression 45°, while its roof is seen at an angle of elevation 30°.

Construction survey: 18 m building viewing another building’s foot and roof.

(a) Find the horizontal distance between the buildings.

(b) Find how far the second building’s roof is above the surveyor’s level.

(d) A 6 m antenna is then installed vertically on its roof. If θ is the new angle of elevation from the original survey point to the antenna top, find tan θ.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Complete Case-Study Solution

Part (a)

Let horizontal distance = d.

Using the foot: tan 45° = 18/d ⇒ d = 18 m.

Horizontal distance = 18 m.

Part (b)

Let the height difference be x.

tan 30° = x/18 ⇒ 1/√3 = x/18.

x = 18/√3 = 6√3 m.

Roof is 6√3 m above the surveyor’s level.

Part (c)

Second building height = 18 + 6√3 m.

Height = 18 + 6√3 m.

Part (d)

Height of antenna top above surveyor level = 6√3 + 6.

Horizontal distance remains 18 m.

tan θ = (6√3 + 6)/18.

Simplify by 6: tan θ = (√3 + 1)/3.

tan θ = (√3 + 1)/3.

60-Second Applications of Trigonometry Revision

Line of sight
Straight line from observer’s eye to viewed point.

Elevation
Angle above the horizontal at the observer.

Depression
Angle below the observer’s horizontal; equals corresponding ground-level elevation.

Main ratio
tan θ = vertical height / horizontal distance.

Standard values
tan30° = 1/√3, tan45° = 1, tan60° = √3.

Observer height
Add eye/instrument height when total height is required.

Two positions
Write one equation per triangle, then connect the distances.

Final check
Exact surd, correct units, physically sensible answer.

Can You Solve These Without Looking at the Strategy Sheet?

These five ungraded problems are separate from the 95 main entries. They are intended as final HOTS/Olympiad challenges.

Question 96

Ungraded Level 5 — Olympiad Challenge Olympiad Two Towers

From the top of a 20 m building, the foot of a vertical tower is seen at 45° depression and the tower top at 45° elevation. Find the tower height. Then find tan θ, where θ is the angle of elevation of the tower top from the foot of the 20 m building.

A 20 m building sees the tower foot at 45° depression and top at 45° elevation.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let horizontal distance between building and tower = d.

From the depression triangle: tan 45° = 20/d ⇒ d = 20 m.

The tower top is also at 45° elevation from the 20 m level, so height difference above that level = d tan45° = 20 m.

Tower height = 20 + 20 = 40 m.

From the foot of the first building: tan θ = 40/20 = 2.

Tower height = 40 m and tan θ = 2.

Question 97

Ungraded Level 5 — Olympiad Challenge Olympiad Multiple Observers

Two points A and B are 10 m apart on the same side of a tower, with B nearer the tower. The angles of elevation are 30° at A and 45° at B. Point C is the midpoint of AB. Find the tower height and obtain an exact expression for tan ∠C.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let distance from B to the tower = x and tower height = h.

At B: tan45° = h/x ⇒ h = x.

At A: distance = x + 10 and tan30° = h/(x + 10).

1/√3 = x/(x + 10) ⇒ x = 5(√3 + 1). Thus h = 5(√3 + 1) m.

C is 5 m farther from the tower than B, so its distance = x + 5 = 5(√3 + 2).

tan ∠C = h/(x + 5) = [5(√3 + 1)]/[5(√3 + 2)].

h = 5(√3 + 1) m and tan ∠C = (√3 + 1)/(√3 + 2).

Question 98

Ungraded Level 5 — Olympiad Challenge Olympiad Elevation + Depression

A 10 m vertical mast stands on level ground near a cliff. From the cliff top, the mast’s foot is seen at 45° depression and its top at 30° depression. Find the cliff height.

Cliff top views the foot and top of a 10 m mast at two depression angles.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let cliff height = H and horizontal separation = d.

From mast foot: tan45° = H/d ⇒ d = H.

Vertical drop from cliff top to mast top = H − 10.

From mast top: tan30° = (H − 10)/d = (H − 10)/H.

1/√3 = (H − 10)/H ⇒ H = √3(H − 10).

H(√3 − 1) = 10√3.

H = 10√3/(√3 − 1) = 5√3(√3 + 1) = 15 + 5√3 m.

Cliff height = 15 + 5√3 m.

Question 99

Ungraded Level 5 — Olympiad Challenge Olympiad Moving Observer

An observer walks 40 m toward a tower. The angle of elevation changes from 30° to 60°. At the midpoint of the 40 m walk, what is tan of the angle of elevation?

A 40 m walk toward a tower from 30° to 60°, with midpoint marked.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let the final distance from tower be x and height be h.

At 60°: h = x√3.

Initial distance = x + 40, and at 30°: h = (x + 40)/√3.

Equating: x√3 = (x + 40)/√3 ⇒ 3x = x + 40 ⇒ x = 20 m.

Thus h = 20√3 m and initial distance = 60 m.

The midpoint of the walk is 20 m from either endpoint, hence 40 m from the tower.

Therefore tan θ_mid = h/40 = 20√3/40 = √3/2.

tan θ at the midpoint = √3/2.

Question 100

Ungraded Level 5 — Olympiad Challenge Olympiad Geometric Constraint

Two vertical poles of heights 10 m and 20 m stand 10√3 m apart on level ground. Find the point P between them from which the angles of elevation of their tops are equal. Also find the common angle.

Hint: Build or read the right triangle first, identify what is known and what is required, then choose the trigonometric ratio that connects those quantities.

Solution

Let P be x m from the 10 m pole. Then it is 10√3 − x m from the 20 m pole.

If the common angle is θ, then tan θ = 10/x = 20/(10√3 − x).

Cross-multiply: 10(10√3 − x) = 20x.

100√3 − 10x = 20x ⇒ 30x = 100√3.

x = 10√3/3 m.

Then tan θ = 10/(10√3/3) = 3/√3 = √3.

Therefore θ = 60°.

P is 10√3/3 m from the 10 m pole (and 20√3/3 m from the 20 m pole); common angle = 60°.

Some Applications of Trigonometry — Heights & Distances

1. Essential Trigonometry Revision

2. Line of Sight, Angle of Elevation & Angle of Depression

3. How to Convert a Word Problem into a Trigonometric Diagram

4. Basic Height-and-Distance Model

5. Choosing the Correct Trigonometric Ratio

6. Observer’s Height

7. Objects Below the Observer: Cliffs, Lighthouses, Terraces & Bridges

8. Two Observation Points on the Same Side

9. Two Observation Points on Opposite Sides

10. Moving Towards or Away from a Tower

11. Two Buildings / Top and Bottom of Another Object

12. Flagpoles, Antennas and Objects on Buildings

13. Shadow Problems

14. Exact Values and Rationalisation

15. Diagram Reading Skills

16. Applications of Trigonometry — Formula & Strategy Sheet

The 7-Step Method

17. Explore Heights and Distances

18. 10 Common Mistakes in Applications of Trigonometry

19. How to Score Full Marks in Applications of Trigonometry

Question Bank

Section A — 30 MCQs (30 × 1 Mark)

Section B — 15 Two-Mark Questions

Section C — 15 Three-Mark Questions

Section D — 20 Four-Mark Questions

Section E — 10 Five-Mark Questions

Section F — 5 Case Studies (4 Sub-questions Each)

Lighthouse Navigation

Part (a)

Part (b)

Part (c)

Part (d)

School Building and Flagstaff

Part (a)

Part (b)

Part (c)

Part (d)

Drone Survey

Part (a)

Part (b)

Part (c)

Part (d)

Fire Lookout Tower

Part (a)

Part (b)

Part (c)

Part (d)

Construction Survey Between Two Buildings

Part (a)

Part (b)

Part (c)

Part (d)

60-Second Applications of Trigonometry Revision

Can You Solve These Without Looking at the Strategy Sheet?

Share this:

Like this:

Leave a ReplyCancel reply

Let’s connect

Join the fun!

Recent posts

Discover more from SK Tuitions