Class 10 NCERT Solutions
Chapter 9: Some Applications of Trigonometry
Master the angles of elevation and depression to solve real-world heights and distances using the logic of trigonometry and our step-by-step methods.
Exercise 9.1
1. 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 $30°$.
Let height of pole $\text{AB} = x$ m and rope $\text{AC} = 20$ m. In $\triangle\text{ABC}$ with $\angle\text{ACB} = 30°$:
$$\sin 30° = \frac{\text{AB}}{\text{AC}} \Rightarrow \frac{1}{2} = \frac{x}{20} \Rightarrow x = 10 \text{ m}$$$\therefore$ Height of the pole $= \mathbf{10}$ m.
2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30°$ 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 broken portion $\text{DC} = x$ m and the standing portion $\text{BC} = y$ m, with $\text{BD} = 8$ m. Height of tree $= x + y$.
In $\triangle\text{BCD}$:
$$\frac{x}{8} = \sec 30° = \frac{2}{\sqrt{3}} \Rightarrow x = \frac{16}{\sqrt{3}}$$ $$\frac{y}{8} = \tan 30° = \frac{1}{\sqrt{3}} \Rightarrow y = \frac{8}{\sqrt{3}}$$ $$\text{Height of tree} = \frac{16}{\sqrt{3}}+\frac{8}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} \text{ m}$$
3. A contractor plans to install two slides for children. For children below 5 years: slide top height 1.5 m, inclined at $30°$. For elder children: height 3 m, inclined at $60°$. What should be the length of the slide in each case?
For children below 5 years: Let length $= x$ m, height $\text{AB} = 1.5$ m, $\angle\text{C} = 30°$.
$$\frac{\text{AB}}{\text{AC}} = \sin 30° \Rightarrow \frac{1.5}{x} = \frac{1}{2} \Rightarrow x = 3 \text{ m}$$For elder children: Let length $= y$ m, height $\text{PQ} = 3$ m, $\angle\text{R} = 60°$.
$$\frac{\text{PQ}}{\text{PR}} = \sin 60° \Rightarrow \frac{3}{y} = \frac{\sqrt{3}}{2} \Rightarrow y = \frac{6}{\sqrt{3}} = 2\sqrt{3} \approx 3.46 \text{ m}$$
4. 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 $30°$. Find the height of the tower.
Let tower $\text{AB} = x$ m and $\text{BC} = 30$ m. In right-angled $\triangle\text{ABC}$ with $\angle\text{ACB} = 30°$:
$$\frac{\text{AB}}{\text{BC}} = \tan 30° \Rightarrow \frac{x}{30} = \frac{1}{\sqrt{3}} \Rightarrow x = \frac{30}{\sqrt{3}} = 10\sqrt{3} \approx 17.32 \text{ m}$$
5. 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 $60°$. Find the length of the string, assuming that there is no slack in the string.
Let the string length $= x$ m and height $\text{PQ} = 60$ m. In $\triangle\text{POQ}$ with $\angle\text{POQ} = 60°$:
$$\frac{\text{OP}}{\text{PQ}} = \operatorname{cosec} 60° \Rightarrow \frac{x}{60} = \frac{2}{\sqrt{3}} \Rightarrow x = \frac{120}{\sqrt{3}} = 40\sqrt{3} \text{ m}$$
6. 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 $30°$ to $60°$ as he walks towards the building. Find the distance he walked towards the building.
Effective height $\text{AB} = 30 – 1.5 = 28.5$ m. Let the closer distance be $y$ m and the walked distance be $x$ m.
At the closer position ($60°$):
$$\frac{y}{\text{AB}} = \cot 60° \Rightarrow y = \frac{28.5}{\sqrt{3}} = 16.45 \text{ m} \quad \cdots(i)$$At the farther position ($30°$):
$$x+y = 28.5\sqrt{3} = 49.36 \text{ m} \quad \cdots(ii)$$Subtracting (i) from (ii): $x = 49.36 – 16.45 = \mathbf{32.91}$ m.
7. 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 $45°$ and $60°$ respectively. Find the height of the tower.
Let tower $\text{AD} = x$ m, building $\text{BA} = 20$ m, and horizontal distance $\text{BC} = y$ m.
From $\triangle\text{ABC}$ ($\angle\text{ACB} = 45°$):
$$\frac{y}{20} = \cot 45° = 1 \Rightarrow y = 20 \text{ m} \quad \cdots(i)$$From $\triangle\text{DBC}$ ($\angle\text{DCB} = 60°$):
$$\frac{x+20}{y} = \tan 60° = \sqrt{3} \Rightarrow x+20 = 20\sqrt{3}$$ $$\Rightarrow x = 20\sqrt{3}-20 = 20(\sqrt{3}-1) \approx 14.64 \text{ m}$$
8. 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 $60°$ and from the same point the angle of elevation of the top of the pedestal is $45°$. Find the height of the pedestal.
Let height of pedestal $\text{AB} = x$ m. From $\triangle\text{OAB}$ ($\angle\text{AOB} = 45°$):
$$\frac{\text{OA}}{x} = \cot 45° = 1 \Rightarrow \text{OA} = x \quad \cdots(i)$$From $\triangle\text{OAC}$ ($\angle\text{AOC} = 60°$, total height $= x + 1.6$):
$$\frac{x}{x+1.6} = \cot 60° = \frac{1}{\sqrt{3}} \Rightarrow \sqrt{3}\,x = x+1.6 \Rightarrow (\sqrt{3}-1)x = 1.6$$ $$\therefore x = \frac{1.6}{\sqrt{3}-1} = \frac{1.6(\sqrt{3}+1)}{2} = 0.8(\sqrt{3}+1) \approx 2.19 \text{ m}$$
9. The angle of elevation of the top of a building from the foot of the tower is $30°$ and the angle of elevation of the top of the tower from the foot of the building is $60°$. If the tower is 50 m high, find the height of the building.
Let building height $= y$ m and $\text{BD} = x$ m. Tower $\text{CD} = 50$ m.
From $\triangle\text{CDB}$ ($\angle\text{CBD} = 60°$):
$$\frac{x}{50} = \cot 60° = \frac{1}{\sqrt{3}} \Rightarrow x = \frac{50}{\sqrt{3}} \quad \cdots(i)$$From $\triangle\text{ABD}$ ($\angle\text{ADB} = 30°$):
$$\frac{y}{x} = \tan 30° = \frac{1}{\sqrt{3}} \Rightarrow y = \frac{x}{\sqrt{3}} = \frac{50}{3} = 16\tfrac{2}{3} \text{ m}$$
10. 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 $60°$ and $30°$, respectively. Find the heights of the poles and the distances of the point from the poles.
Let each pole height $= h$ m, $\text{BE} = x$ m, and $\text{ED} = (80-x)$ m.
From $\triangle\text{ABE}$ ($\angle\text{AEB} = 60°$): $h = \sqrt{3}\,x \quad \cdots(i)$
From $\triangle\text{CDE}$ ($\angle\text{CED} = 30°$): $h = \dfrac{80-x}{\sqrt{3}} \quad \cdots(ii)$
From (i) and (ii): $\sqrt{3}\,x = \dfrac{80-x}{\sqrt{3}} \Rightarrow 3x = 80-x \Rightarrow x = 20$ m
$\therefore$ Distance from one pole $= 20$ m, distance from the other $= 60$ m.
Height of each pole: $h = 20\sqrt{3}$ m.
11. 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 $60°$. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $30°$. Find the height of the tower and the width of the canal.
Let width $\text{BC} = x$ m and tower height $\text{AB} = h$ m.
From $\triangle\text{ABC}$ ($60°$): $h = \sqrt{3}\,x \quad \cdots(i)$
From $\triangle\text{ABD}$ ($30°$, $\text{BD} = 20+x$): $h = \dfrac{20+x}{\sqrt{3}} \quad \cdots(ii)$
From (i) and (ii): $3x = 20+x \Rightarrow 2x = 20 \Rightarrow x = 10$ m
$\therefore$ Width of canal $= \mathbf{10}$ m and height of tower $= 10\sqrt{3}$ m.
12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $60°$ and the angle of depression of its foot is $45°$. Determine the height of the tower.
Let $\text{AB} = 7$ m (building), $\text{AR} = \text{BQ} = x$ m (horizontal), and $\text{PR} = y$ m (height above building level).
From $\triangle\text{ABQ}$ ($\angle\text{RAQ} = 45°$): $\dfrac{7}{x} = \tan 45° = 1 \Rightarrow x = 7$ m
From $\triangle\text{PRA}$ ($\angle\text{PAR} = 60°$): $\dfrac{y}{x} = \tan 60° = \sqrt{3} \Rightarrow y = 7\sqrt{3}$ m
$$\text{Height of tower} = 7 + y = 7+7\sqrt{3} = 7(1+\sqrt{3}) = 7\times2.732 \approx 19.12 \text{ m}$$
13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are $30°$ and $45°$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Lighthouse $\text{AB} = 75$ m. Let $\text{BC} = x$ m (nearer ship) and $\text{BD} = y$ m (farther ship).
From $\triangle\text{ABC}$ ($45°$): $\dfrac{x}{75} = \cot 45° = 1 \Rightarrow x = 75$ m
From $\triangle\text{ABD}$ ($30°$): $\dfrac{y}{75} = \cot 30° = \sqrt{3} \Rightarrow y = 75\sqrt{3}$ m
$$\text{Distance between two ships} = y-x = 75\sqrt{3}-75 = 75(\sqrt{3}-1) \text{ m}$$
14. 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 $60°$. After some time, the angle of elevation reduces to $30°$. Find the distance travelled by the balloon during the interval.
Effective balloon height $= 88.2 – 1.2 = 87$ m above the girl’s eyes.
At the first position P ($60°$):
$$\text{AC} = \frac{87}{\tan 60°} = \frac{87}{\sqrt{3}} \quad \cdots(i)$$At the second position R ($30°$):
$$\text{AD} = \frac{87}{\tan 30°} = 87\sqrt{3} \quad \cdots(ii)$$ $$\text{Distance CD} = \text{AD}-\text{AC} = 87\sqrt{3}-\frac{87}{\sqrt{3}} = 87\cdot\frac{3-1}{\sqrt{3}} = \frac{87\times2}{\sqrt{3}} = 58\sqrt{3} \text{ m}$$
15. 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 $30°$, 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 $60°$. Find the time taken by the car to reach the foot of the tower from this point.
Let tower $\text{AB} = h$ m, $\text{DC} = x$ m (distance covered in 6s), and $\text{BC} = y$ m (remaining distance).
When car is at D ($30°$): $x+y = \sqrt{3}\,h \quad \cdots(i)$
When car is at C ($60°$): $y = \dfrac{h}{\sqrt{3}} \Rightarrow h = \sqrt{3}\,y \quad \cdots(ii)$
Substituting (ii) in (i): $x+y = 3y \Rightarrow x = 2y \Rightarrow y = \dfrac{x}{2}$
Since distance $x$ is covered in 6 seconds, distance $y = \dfrac{x}{2}$ will be covered in 3 seconds.
16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Let the angles of elevation from C (4 m away) and D (9 m away) be $\alpha$ and $(90°-\alpha)$ respectively.
From $\triangle\text{ABC}$: $\dfrac{h}{4} = \tan\alpha \quad \cdots(i)$
From $\triangle\text{ABD}$: $\dfrac{h}{9} = \tan(90°-\alpha) = \cot\alpha \quad \cdots(ii)$
Multiplying (i) and (ii):
$$\frac{h}{4}\cdot\frac{h}{9} = \tan\alpha\cdot\cot\alpha = 1 \Rightarrow h^2 = 36 \Rightarrow h = 6 \text{ m} \qquad \blacksquare$$