NCERT Class 10 Maths – Exercise 9.1 Solutions

NCERT Class 10 Maths

Chapter 9 – Some Applications of Trigonometry | Exercise 9.1

(Rationalized Syllabus 2025-26)

💡 Important Ratios

  • $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$
  • $\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}$
  • $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$
Q1

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^\circ$.

Let $AB$ be the pole and $AC$ be the rope.
In right $\triangle ABC$:
Length of rope (Hypotenuse) $AC = 20$ m. Angle $C = 30^\circ$.
$$\sin 30^\circ = \frac{AB}{AC}$$
$$\frac{1}{2} = \frac{AB}{20} \Rightarrow AB = \frac{20}{2} = 10$$
✔️ Height of the pole is 10 m.
Q2

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^\circ$ with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Let the tree be $AB$, broken at point $C$. The top $A$ touches the ground at $D$.
In right $\triangle CBD$: $BD = 8$ m, $\angle D = 30^\circ$.
Find broken part (Hypotenuse CD):
$$\cos 30^\circ = \frac{BD}{CD} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{CD} \Rightarrow CD = \frac{16}{\sqrt{3}}$$
Find standing part (Perpendicular BC):
$$\tan 30^\circ = \frac{BC}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{8} \Rightarrow BC = \frac{8}{\sqrt{3}}$$
Total height $= BC + CD = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}}$.
Rationalizing: $\frac{24\sqrt{3}}{3} = 8\sqrt{3}$.
✔️ Height of the tree is $8\sqrt{3}$ m.
Q3

A contractor plans to install two slides for children. For children below 5 years: height 1.5 m, inclined at $30^\circ$. For elder children: height 3 m, inclined at $60^\circ$. What should be the length of the slide in each case?

Case 1: Below 5 years
Height (Perp) $= 1.5$ m, Angle $= 30^\circ$. Let slide length be $l_1$.
$$\sin 30^\circ = \frac{1.5}{l_1} \Rightarrow \frac{1}{2} = \frac{1.5}{l_1} \Rightarrow l_1 = 3$$
✔️ Length of slide = 3 m
Case 2: Elder Children
Height (Perp) $= 3$ m, Angle $= 60^\circ$. Let slide length be $l_2$.
$$\sin 60^\circ = \frac{3}{l_2} \Rightarrow \frac{\sqrt{3}}{2} = \frac{3}{l_2} \Rightarrow l_2 = \frac{6}{\sqrt{3}} = 2\sqrt{3}$$
✔️ Length of slide = $2\sqrt{3}$ m
Q4

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is $30^\circ$. Find the height of the tower.

Let height of tower be $h$. Base distance $= 30$ m. Angle $= 30^\circ$.
$$\tan 30^\circ = \frac{h}{30}$$
$$\frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10\sqrt{3}$$
✔️ Height of tower is $10\sqrt{3}$ m.
Q5

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^\circ$. Find the length of the string, assuming that there is no slack in the string.

Height (Perp) $= 60$ m. Angle $= 60^\circ$. Let string length be $x$.
$$\sin 60^\circ = \frac{60}{x}$$
$$\frac{\sqrt{3}}{2} = \frac{60}{x} \Rightarrow x = \frac{120}{\sqrt{3}} = 40\sqrt{3}$$
✔️ Length of string is $40\sqrt{3}$ m.
Q6

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from $30^\circ$ to $60^\circ$ as he walks towards the building. Find the distance he walked towards the building.

Effective height of building above boy’s eye level $= 30 – 1.5 = 28.5$ m.
Let distances from building be $x$ (at $30^\circ$) and $y$ (at $60^\circ$). Distance walked $= x – y$.
At $60^\circ$:
$$\tan 60^\circ = \frac{28.5}{y} \Rightarrow \sqrt{3} = \frac{28.5}{y} \Rightarrow y = \frac{28.5}{\sqrt{3}}$$
At $30^\circ$:
$$\tan 30^\circ = \frac{28.5}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{x} \Rightarrow x = 28.5\sqrt{3}$$
Distance walked:
$$x – y = 28.5\sqrt{3} – \frac{28.5}{\sqrt{3}} = 28.5 \left(\sqrt{3} – \frac{1}{\sqrt{3}}\right)$$
$$= 28.5 \left(\frac{3-1}{\sqrt{3}}\right) = \frac{57}{\sqrt{3}} = 19\sqrt{3}$$
✔️ Distance walked is $19\sqrt{3}$ m.
Q7

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.

Let building height $BC = 20$ m and tower height $AB = h$.
Observation point $D$ on ground. $\angle BDC = 45^\circ$, $\angle ADC = 60^\circ$.
In $\triangle BCD$:
$$\tan 45^\circ = \frac{20}{CD} \Rightarrow 1 = \frac{20}{CD} \Rightarrow CD = 20$$
In $\triangle ACD$: Total height $= 20 + h$.
$$\tan 60^\circ = \frac{20 + h}{CD} \Rightarrow \sqrt{3} = \frac{20 + h}{20}$$
$$20\sqrt{3} = 20 + h \Rightarrow h = 20(\sqrt{3} – 1)$$
✔️ Height of tower is $20(\sqrt{3} – 1)$ m.
Q8

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ$. Find the height of the pedestal.

Let pedestal height be $h$. Statue height $= 1.6$ m. Base distance $= x$.
For pedestal (Top is $45^\circ$):
$$\tan 45^\circ = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h$$
For statue top (Top is $60^\circ$): Total height $= h + 1.6$.
$$\tan 60^\circ = \frac{h + 1.6}{x} \Rightarrow \sqrt{3} = \frac{h + 1.6}{h}$$
$$h\sqrt{3} = h + 1.6 \Rightarrow h(\sqrt{3} – 1) = 1.6$$
$$h = \frac{1.6}{\sqrt{3} – 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1)$$
✔️ Height of pedestal is $0.8(\sqrt{3} + 1)$ m.
Q9

The angle of elevation of the top of a building from the foot of the tower is $30^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^\circ$. If the tower is 50 m high, find the height of the building.

Tower $AB = 50$ m. Building $CD = h$. Distance between them $= x$.
From foot of building to top of tower ($60^\circ$):
$$\tan 60^\circ = \frac{50}{x} \Rightarrow \sqrt{3} = \frac{50}{x} \Rightarrow x = \frac{50}{\sqrt{3}}$$
From foot of tower to top of building ($30^\circ$):
$$\tan 30^\circ = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{50/\sqrt{3}}$$
$$h = \frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{50}{3} = 16\frac{2}{3}$$
✔️ Height of building is $16\frac{2}{3}$ m.
Q10

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^\circ$ and $30^\circ$, respectively. Find the height of the poles and the distances of the point from the poles.

Let height of poles be $h$. Let point be distance $x$ from first pole, so $(80 – x)$ from second.
Pole 1 ($60^\circ$): $\tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3}$.
Pole 2 ($30^\circ$): $\tan 30^\circ = \frac{h}{80 – x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{80 – x}$.
$$80 – x = 3x \Rightarrow 4x = 80 \Rightarrow x = 20$$
Second distance $= 80 – 20 = 60$.
Height $h = 20\sqrt{3}$.
✔️ Height: $20\sqrt{3}$ m. Distances: 20 m and 60 m.
Q11

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^\circ$. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is $30^\circ$. Find the height of the tower and the width of the canal.

Let tower height be $h$ and canal width be $x$.
At bank ($60^\circ$): $\tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3}$.
20 m away ($30^\circ$): Base becomes $x + 20$.
$$\tan 30^\circ = \frac{h}{x + 20} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 20}$$
$$x + 20 = 3x \Rightarrow 2x = 20 \Rightarrow x = 10$$
Height $h = 10\sqrt{3}$.
✔️ Height: $10\sqrt{3}$ m. Width: 10 m.
Q12

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Determine the height of the tower.

Building $AB = 7$ m. Tower $CD = H$. Let horizontal distance be $x$.
Depression of foot ($45^\circ$):
$$\tan 45^\circ = \frac{7}{x} \Rightarrow x = 7$$
Elevation of top ($60^\circ$): Height above building level is $h’$.
$$\tan 60^\circ = \frac{h’}{7} \Rightarrow h’ = 7\sqrt{3}$$
Total height $H = 7 + 7\sqrt{3} = 7(\sqrt{3} + 1)$.
✔️ Height of tower is $7(\sqrt{3} + 1)$ m.
Q13

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are $30^\circ$ and $45^\circ$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Height $h = 75$ m. Let distance of first ship ($45^\circ$) be $y$ and second ($30^\circ$) be $x$.
Ship 1 ($45^\circ$): $\tan 45^\circ = \frac{75}{y} \Rightarrow y = 75$.
Ship 2 ($30^\circ$): $\tan 30^\circ = \frac{75}{x} \Rightarrow x = 75\sqrt{3}$.
Distance between ships $= x – y = 75\sqrt{3} – 75$.
✔️ Distance is $75(\sqrt{3} – 1)$ m.
Q14

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60^\circ$. After some time, the angle of elevation reduces to $30^\circ$. Find the distance travelled by the balloon during the interval.

Effective height $= 88.2 – 1.2 = 87$ m.
First position ($60^\circ$): Distance $x$.
$$\tan 60^\circ = \frac{87}{x} \Rightarrow x = \frac{87}{\sqrt{3}} = 29\sqrt{3}$$
Second position ($30^\circ$): Distance $y$.
$$\tan 30^\circ = \frac{87}{y} \Rightarrow y = 87\sqrt{3}$$
Distance travelled $= y – x = 87\sqrt{3} – 29\sqrt{3} = 58\sqrt{3}$.
✔️ Distance travelled is $58\sqrt{3}$ m.
Q15

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^\circ$, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be $60^\circ$. Find the time taken by the car to reach the foot of the tower from this point.

Let height be $h$.
At $30^\circ$, distance $d_1 = h\sqrt{3}$.
At $60^\circ$, distance $d_2 = h/\sqrt{3}$.
Distance travelled in 6 sec $= d_1 – d_2 = h\sqrt{3} – \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$.
Speed $v = \frac{\text{Distance}}{\text{Time}} = \frac{2h}{6\sqrt{3}} = \frac{h}{3\sqrt{3}}$.
Time to cover remaining $d_2$:
$$t = \frac{d_2}{v} = \frac{h/\sqrt{3}}{h/3\sqrt{3}} = 3$$
✔️ Time taken is 3 seconds.
🎉 Exercise 9.1 Completed | Chapter 9 Finished!
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