NCERT Mathematics · Class 10 · English · 2026-27

Some Applications of Trigonometry — Exercise 9.1

Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.

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Question 1

A taut 20 m rope makes 30° with level ground. Find the vertical pole height.

Step-by-step solution

The rope is the hypotenuse, and the height h is opposite 30°. Thus sin30°=h/20, giving h=20×1/2.

Right triangle model, not to scalehground20 m30°
Right triangle model; not to scale.
Answer: 10 m.

Resource ID: CB-10-M-CH9-EX9.1-Q1

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Question 2

A broken tree touches the ground 8 m from its base, making 30° with the ground. Find its original height.

Step-by-step solution

Let the standing stump have height h and the broken sloping part length L. tan30°=h/8 gives h=8/√3. Also cos30°=8/L gives L=16/√3. The original height is h+L=24/√3.

Right triangle model, not to scaleh8 mL30°
Right triangle model; not to scale.
Answer: 8√3 m.

Resource ID: CB-10-M-CH9-EX9.1-Q2

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Question 3

Find slide lengths for heights/angles (1.5 m,30°) and (3 m,60°).

Step-by-step solution

The slide is the hypotenuse, so L=h/sinθ. First: L=1.5/(1/2)=3 m. Second: L=3/(√3/2)=6/√3.

Right triangle model, not to scalehgroundLθ
Right triangle model; not to scale.
Answer: Younger children: 3 m; older children: 2√3 m.

Resource ID: CB-10-M-CH9-EX9.1-Q3

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Question 4

From 30 m away, the top of a tower has elevation 30°. Find its height.

Step-by-step solution

With horizontal distance 30 m, tan30°=h/30. Therefore h=30/√3.

Right triangle model, not to scaleh30 msight line30°
Right triangle model; not to scale.
Answer: 10√3 m.

Resource ID: CB-10-M-CH9-EX9.1-Q4

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Question 5

A kite is 60 m high, with its taut string tied at ground level at 60°. Find the string length.

Step-by-step solution

The string length L is the hypotenuse. sin60°=60/L, so L=60×2/√3.

Right triangle model, not to scale60 mgroundL60°
Right triangle model; not to scale.
Answer: 40√3 m.

Resource ID: CB-10-M-CH9-EX9.1-Q5

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Question 6

A boy of eye height 1.5 m walks towards a 30 m building as elevation changes from 30° to 60°. Find the distance walked.

Step-by-step solution

Use the height above eye level H=30−1.5=28.5 m. Let the farther and nearer distances be x and y. tan30°=H/x gives x=28.5√3; tan60°=H/y gives y=28.5/√3=9.5√3. The walk is x−y.

Answer: 19√3 m.

Resource ID: CB-10-M-CH9-EX9.1-Q6

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Question 7

A tower sits on a 20 m building; elevation to its base is 45° and to its top 60°. Find tower height.

Step-by-step solution

Let observation distance be x and tower height t. tan45°=20/x gives x=20. Then tan60°=(20+t)/20, so 20+t=20√3. Subtract the building height.

Answer: 20(√3−1) m.

Resource ID: CB-10-M-CH9-EX9.1-Q7

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Question 8

A 1.6 m statue stands on a pedestal. Elevations to the pedestal and statue tops are 45° and 60°. Find pedestal height.

Step-by-step solution

Let pedestal height be h and horizontal distance x. tan45°=h/x gives x=h. Then tan60°=(h+1.6)/h. Hence (√3−1)h=1.6 and h=1.6/(√3−1)=0.8(√3+1).

Answer: 0.8(√3+1) m.

Resource ID: CB-10-M-CH9-EX9.1-Q8

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Question 9

A 50 m tower and a building are observed from each other’s feet at elevations 60° and 30°. Find building height.

Step-by-step solution

Let separation be x. The tower gives tan60°=50/x, so x=50/√3. For building height h, tan30°=h/x; hence h=x/√3=50/3.

Answer: 50/3 m.

Resource ID: CB-10-M-CH9-EX9.1-Q9

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Question 10

Equal poles stand across an 80 m road. A point between them sees elevations 60° and 30°. Find heights and distances.

Step-by-step solution

Let the distance to the 60° pole be x, so the other distance is 80−x. Equal height h gives h=x√3=(80−x)/√3. Multiply by √3: 3x=80−x, giving x=20. Thus the other distance is 60 and h=20√3.

Answer: Each pole is 20√3 m high; distances are 20 m (60° side) and 60 m (30° side).

Resource ID: CB-10-M-CH9-EX9.1-Q10

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Question 11

Across a canal the tower elevation is 60°; from 20 m farther back it is 30°. Find tower height and canal width.

Step-by-step solution

Let canal width be w and height h. Then h=w√3 and h=(w+20)/√3. Equate: 3w=w+20, hence w=10 and h=10√3. The second observation is farther from the tower, so its horizontal distance is w+20.

Answer: Width 10 m; tower height 10√3 m.

Resource ID: CB-10-M-CH9-EX9.1-Q11

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Question 12

From a 7 m building roof, tower-top elevation is 60° and tower-foot depression 45°. Find tower height.

Step-by-step solution

The depression angle equals the elevation angle from the tower foot. If separation is x, tan45°=7/x gives x=7. The height above the roof is x tan60°=7√3. Add the roof height 7.

Answer: 7(1+√3) m.

Resource ID: CB-10-M-CH9-EX9.1-Q12

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Question 13

From a 75 m lighthouse, two ships on the same side have depressions 30° and 45°. Find their separation.

Step-by-step solution

The 30° ship is farther: x=75/tan30°=75√3. The 45° ship has distance y=75. Because they lie on the same ray, their separation is x−y rather than their sum.

Answer: 75(√3−1) m.

Resource ID: CB-10-M-CH9-EX9.1-Q13

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Question 14

A balloon at 88.2 m moves horizontally away from a girl of eye height 1.2 m; elevation drops from 60° to 30°. Find travel distance.

Step-by-step solution

The height above the eyes stays H=88.2−1.2=87 m. The initial horizontal distance is 87/√3=29√3; final distance is 87√3. Subtract initial from final.

Answer: 58√3 m.

Resource ID: CB-10-M-CH9-EX9.1-Q14

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Question 15

A car approaching a tower changes depression from 30° to 60° in six seconds at constant speed. How much time remains?

Step-by-step solution

Let tower height be h. The car’s first and second distances are x=h√3 and y=h/√3. In six seconds it travels x−y=2h/√3=2y. Thus its speed is 2y/6=y/3 per second. Remaining time=y/(y/3).

Answer: 3 seconds from the second observation.

Resource ID: CB-10-M-CH9-EX9.1-Q15

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