Skip to main content
NCERT Solutions · Class 10 · Chapter 9

NCERT Solutions for Class 10 Maths Chapter 9: Some Applications of Trigonometry

Heights and distances: angles of elevation and depression, solved with right triangles and \(\tan\), \(\sin\) or \(\cos\) of \(30^\circ, 45^\circ, 60^\circ\). Our own step-by-step solutions to every question in Exercise 9.1, set out for step marks.

  • 16 questions, 16 parts
  • Every answer checked by computer algebra
  • Free, no sign-in

Try each question first, then open its solution.

Exercise 9.1: Heights and distances

What it tests. Draw the figure, mark the right angle and the given angle, and pick the ratio linking the known side to the unknown (usually \(\tan\)). With two angles, write two equations in the same unknowns. An angle of depression from the top equals the angle of elevation from the bottom (alternate angles).

Exercise 9.1, Question 1

A circus artist climbs a 20 m rope stretched from the top of a vertical pole to the ground, making \(30^\circ\) with the ground. Find the height of the pole.
Show solution
  1. The rope is the hypotenuse: \(\sin 30^\circ = \dfrac{h}{20}\).
  2. \(h = 20 \times \tfrac12 = 10\).
Answer: \(10\) m

Practise this: Step 1, Secure the basics →

Exercise 9.1, Question 2

A tree breaks; the top part bends to touch the ground \(8\) m from the foot, making \(30^\circ\) with the ground. Find the original height.
Show solution
  1. Standing part \(AB\), broken part \(AC\) (hypotenuse), \(BC = 8\).
  2. \(AB = 8\tan 30^\circ = \dfrac{8}{\sqrt3}\); \[\begin{aligned}AC &= \dfrac{8}{\cos 30^\circ} \\ &= \dfrac{16}{\sqrt3}\end{aligned}\]
  3. Height \(= \dfrac{24}{\sqrt3} = 8\sqrt3\).
Answer: \(8\sqrt3\) m (about 13.86 m)

Practise this: Step 2, Board standard →

Exercise 9.1, Question 3

A slide for young children has its top 1.5 m high and is inclined at \(30^\circ\) to the ground; one for older children is 3 m high at \(60^\circ\). Find the length of each slide.
Show solution
  1. Slide length is the hypotenuse: \(L = \dfrac{h}{\sin\theta}\).
  2. \(\dfrac{1.5}{\sin 30^\circ} = 3\); \[\begin{aligned}\dfrac{3}{\sin 60^\circ} &= \dfrac{6}{\sqrt3} \\ &= 2\sqrt3\end{aligned}\]
Answer: \(3\) m and \(2\sqrt3\) m

Practise this: Step 1, Secure the basics →

Exercise 9.1, Question 4

The angle of elevation of the top of a tower from a point 30 m from its foot is \(30^\circ\). Find its height.
Show solution
  1. \[\begin{aligned}&\tan 30^\circ = \dfrac{h}{30} \\ \Rightarrow\ &h = \dfrac{30}{\sqrt3}\end{aligned}\]
Answer: \(10\sqrt3\) m

Practise this: Step 1, Secure the basics →

Exercise 9.1, Question 5

A kite is 60 m above the ground and its string makes \(60^\circ\) with the ground. Find the length of the string (assume it is straight).
Show solution
  1. \[\begin{aligned}&\sin 60^\circ = \dfrac{60}{L} \\ \Rightarrow\ &L = \dfrac{120}{\sqrt3}\end{aligned}\]
Answer: \(40\sqrt3\) m

Practise this: Step 1, Secure the basics →

Exercise 9.1, Question 6

A boy 1.5 m tall stands near a 30 m building. As he walks towards it, the angle of elevation of its top from his eyes rises from \(30^\circ\) to \(60^\circ\). How far did he walk?
Show solution
  1. Height above eye level: \(30 - 1.5 = 28.5\) m.
  2. Distances: \(\dfrac{28.5}{\tan 30^\circ} = 28.5\sqrt3\) and \[\dfrac{28.5}{\tan 60^\circ} = \dfrac{28.5}{\sqrt3}\]
  3. Walked \[\begin{aligned}= 28.5\left(\sqrt3 - \dfrac{1}{\sqrt3}\right) &= 28.5 \times \dfrac{2}{\sqrt3} \\ &= 19\sqrt3\end{aligned}\]
Answer: \(19\sqrt3\) m

Practise this: Step 2, Board standard →

Exercise 9.1, Question 7

From a point on the ground, the angles of elevation of the bottom and top of a transmission tower on a 20 m building are \(45^\circ\) and \(60^\circ\). Find the tower's height.
Show solution
  1. Bottom: \[\begin{aligned}&\tan 45^\circ = \dfrac{20}{x} \\ \Rightarrow\ &x = 20\end{aligned}\]
  2. Top: \[\begin{aligned}&\tan 60^\circ = \dfrac{20 + h}{20} \\ \Rightarrow\ &h = 20\sqrt3 - 20\end{aligned}\]
Answer: \(20(\sqrt3 - 1)\) m

Practise this: Step 2, Board standard →

Exercise 9.1, Question 8

A 1.6 m statue stands on a pedestal. From a point on the ground the angles of elevation of the top of the statue and the top of the pedestal are \(60^\circ\) and \(45^\circ\). Find the height of the pedestal.
Show solution
  1. Pedestal \(h\), distance \(x\): \[\begin{aligned}&\tan 45^\circ = \dfrac hx \\ \Rightarrow\ &x = h\end{aligned}\]
  2. \[\begin{aligned}&\tan 60^\circ = \dfrac{h + 1.6}{h} \\ \Rightarrow\ &h\sqrt3 = h + 1.6\end{aligned}\]
  3. \[\begin{aligned}h &= \dfrac{1.6}{\sqrt3 - 1} \\ &= 0.8(\sqrt3 + 1)\end{aligned}\]
Answer: \(0.8(\sqrt3 + 1)\) m (about 2.19 m)

Practise this: Step 2, Board standard →

Exercise 9.1, Question 9

The angle of elevation of the top of a building from the foot of a 50 m tower is \(30^\circ\), and of the top of the tower from the foot of the building is \(60^\circ\). Find the height of the building.
Show solution
  1. Tower: \[\begin{aligned}&\tan 60^\circ = \dfrac{50}{d} \\ \Rightarrow\ &d = \dfrac{50}{\sqrt3}\end{aligned}\]
  2. Building: \[\begin{aligned}h &= d\tan 30^\circ \\ &= \dfrac{50}{\sqrt3} \times \dfrac{1}{\sqrt3} \\ &= \dfrac{50}{3}\end{aligned}\]
Answer: \(16\tfrac23\) m

Practise this: Step 2, Board standard →

Exercise 9.1, Question 10

Two equal poles stand on opposite sides of an 80 m road. From a point on the road between them, the angles of elevation of their tops are \(60^\circ\) and \(30^\circ\). Find the height of the poles and the distances of the point from them.
Show solution
  1. \(h = x\tan 60^\circ = x\sqrt3\) and \[\begin{aligned}h &= (80 - x)\tan 30^\circ \\ &= \dfrac{80 - x}{\sqrt3}\end{aligned}\]
  2. \(3x = 80 - x \Rightarrow x = 20\); \(h = 20\sqrt3\).
Answer: Height \(20\sqrt3\) m; 20 m from one pole and 60 m from the other

Practise this: Step 3, Full marks on long answers →

Exercise 9.1, Question 11

From a point on one bank of a canal directly opposite a TV tower, the angle of elevation of its top is \(60^\circ\). From a point 20 m further away in line, it is \(30^\circ\). Find the tower's height and the canal's width.
Show solution
  1. \(h = x\sqrt3\) and \(h = \dfrac{x + 20}{\sqrt3}\).
  2. \(3x = x + 20 \Rightarrow x = 10\), \(h = 10\sqrt3\).
Answer: Height \(10\sqrt3\) m, width 10 m

Practise this: Step 3, Full marks on long answers →

Exercise 9.1, Question 12

From the top of a 7 m building, the angle of elevation of the top of a cable tower is \(60^\circ\) and the angle of depression of its foot is \(45^\circ\). Find the tower's height.
Show solution
  1. Depression \(45^\circ\): horizontal distance \(= 7\) m.
  2. Above the building's top: \(7\tan 60^\circ = 7\sqrt3\).
  3. Tower \(= 7 + 7\sqrt3\).
Answer: \(7(\sqrt3 + 1)\) m

Practise this: Step 3, Full marks on long answers →

Exercise 9.1, Question 13

From the top of a 75 m lighthouse, the angles of depression of two ships (one behind the other, on the same side) are \(30^\circ\) and \(45^\circ\). How far apart are the ships?
Show solution
  1. Nearer ship: \(\dfrac{75}{\tan 45^\circ} = 75\). Farther ship: \(\dfrac{75}{\tan 30^\circ} = 75\sqrt3\).
  2. Distance \(= 75\sqrt3 - 75\).
Answer: \(75(\sqrt3 - 1)\) m

Practise this: Step 3, Full marks on long answers →

Exercise 9.1, Question 14

A girl 1.2 m tall watches a balloon moving horizontally at 88.2 m above the ground. The angle of elevation from her eyes changes from \(60^\circ\) to \(30^\circ\). How far did the balloon travel?
Show solution
  1. Height above eye level: \(87\) m.
  2. Distances: \[\dfrac{87}{\tan 60^\circ} = \dfrac{87}{\sqrt3}\] and \(\dfrac{87}{\tan 30^\circ} = 87\sqrt3\).
  3. Travelled \[= 87\left(\sqrt3 - \dfrac1{\sqrt3}\right) = 58\sqrt3\]
Answer: \(58\sqrt3\) m

Practise this: Step 3, Full marks on long answers →

Exercise 9.1, Question 15

A man on top of a tower watches a car approaching at uniform speed. The angle of depression changes from \(30^\circ\) to \(60^\circ\) in 6 seconds. How much longer until the car reaches the tower?
Show solution
  1. Tower height \(h\): distances \(h\sqrt3\) and \(\dfrac{h}{\sqrt3}\).
  2. In 6 s it covers \[h\sqrt3 - \dfrac{h}{\sqrt3} = \dfrac{2h}{\sqrt3}\]; it still has \(\dfrac{h}{\sqrt3}\) to go, half as far.
  3. So it takes half the time: 3 s.
Answer: 3 seconds

Practise this: Step 4, 95+ stretch (HOTS) →

Exercise 9.1, Question 16

The angles of elevation of the top of a tower from two points 4 m and 9 m from its base, on the same side and in line with it, are complementary. Prove the height is 6 m.
Show solution
  1. Let the angles be \(\theta\) and \(90^\circ - \theta\): \(\tan\theta = \dfrac h4\) and \[\begin{aligned}\tan(90^\circ - \theta) &= \cot\theta \\ &= \dfrac h9\end{aligned}\]
  2. Multiplying, \[\begin{aligned}\tan\theta\cot\theta &= 1 \\ &= \dfrac{h^2}{36}\end{aligned}\], so \(h = 6\).
Answer: Proved: \(h = 6\) m

Where marks slip: Reject \(h = -6\): a height is positive.

Practise this: Step 4, 95+ stretch (HOTS) →

Done the NCERT exercises? The board paper asks more

Some Applications of Trigonometry has 37 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

Some Applications of Trigonometry: notes and questionsCreate a free accountTeachers: set it as homework

Also useful: free MCQs and case studies for Some Applications of Trigonometry · Class 10 formula sheet · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.