NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1#
Exercise 9.1 applies trigonometry to real-life problems involving heights and distances. Students use angles of elevation and depression to find unknown heights and distances.
Key Concepts:
- Angle of Elevation: The angle formed between the horizontal line and the line of sight when looking upward at an object.
- Angle of Depression: The angle formed between the horizontal line and the line of sight when looking downward at an object.
- Standard values: $\tan 30° = \frac{1}{\sqrt{3}}$, $\tan 45° = 1$, $\tan 60° = \sqrt{3}$
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°. #
Answer
Let AB = height of pole, AC = rope = 20 m, angle at C = 30°.
$$\sin 30° = \frac{AB}{AC}$$$$\frac{1}{2} = \frac{AB}{20}$$$$AB = 10 \text{ m}$$∴ The height of the pole is 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. #
Answer
Let BC = remaining part of tree, AB = broken part (now hypotenuse), AC = 8 m.
$\tan 30° = \dfrac{BC}{AC}$
$\dfrac{1}{\sqrt{3}} = \dfrac{BC}{8} \Rightarrow BC = \dfrac{8}{\sqrt{3}} = \dfrac{8\sqrt{3}}{3}$ m
$\cos 30° = \dfrac{AC}{AB}$
$\dfrac{\sqrt{3}}{2} = \dfrac{8}{AB} \Rightarrow AB = \dfrac{16}{\sqrt{3}} = \dfrac{16\sqrt{3}}{3}$ m
Height of tree = $BC + AB = \dfrac{8\sqrt{3}}{3} + \dfrac{16\sqrt{3}}{3} = \dfrac{24\sqrt{3}}{3} = 8\sqrt{3}$ m
∴ Height of the tree = $8\sqrt{3} \approx$ 13.86 m.
3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? #
Answer
For younger children (30°, height = 1.5 m):
$$\sin 30° = \frac{1.5}{l_1}$$$$\frac{1}{2} = \frac{1.5}{l_1} \Rightarrow l_1 = 3 \text{ m}$$For older children (60°, height = 3 m):
$$\sin 60° = \frac{3}{l_2}$$$$\frac{\sqrt{3}}{2} = \frac{3}{l_2} \Rightarrow l_2 = \frac{6}{\sqrt{3}} = 2\sqrt{3} \text{ m}$$∴ Length of slide for younger children = 3 m, for older children = $2\sqrt{3}$ m ≈ 3.46 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. #
Answer
Let h = height of tower, distance = 30 m.
$$\tan 30° = \frac{h}{30}$$$$\frac{1}{\sqrt{3}} = \frac{h}{30}$$$$h = \frac{30}{\sqrt{3}} = 10\sqrt{3} \text{ m}$$∴ Height of tower = $10\sqrt{3} \approx$ 17.32 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. #
Answer
∴ Length of string = $40\sqrt{3} \approx$ 69.28 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. #
Answer
Height of building above boy’s eyes = $30 - 1.5 = 28.5$ m.
Let initial distance = $d_1$, final distance = $d_2$.
From angle 30°: $\tan 30° = \dfrac{28.5}{d_1} \Rightarrow d_1 = 28.5\sqrt{3}$ m
From angle 60°: $\tan 60° = \dfrac{28.5}{d_2} \Rightarrow d_2 = \dfrac{28.5}{\sqrt{3}} = \dfrac{28.5\sqrt{3}}{3} = 9.5\sqrt{3}$ m
Distance walked = $d_1 - d_2 = 28.5\sqrt{3} - 9.5\sqrt{3} = 19\sqrt{3}$ m
∴ Distance walked = $19\sqrt{3} \approx$ 32.9 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. #
Answer
Let the horizontal distance = d, height of tower = h.
From angle 45° (top of building):
$$\tan 45° = \frac{20}{d} \Rightarrow d = 20 \text{ m}$$From angle 60° (top of tower):
$$\tan 60° = \frac{20+h}{d} = \frac{20+h}{20}$$$$\sqrt{3} = \frac{20+h}{20}$$$$20+h = 20\sqrt{3}$$$$h = 20(\sqrt{3}-1) \text{ m}$$∴ Height of tower = $20(\sqrt{3}-1) \approx$ 14.64 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. #
Answer
Let h = height of pedestal, d = horizontal distance.
Angle to top of pedestal = 45°: $\tan 45° = \dfrac{h}{d} \Rightarrow d = h$
Angle to top of statue = 60°: $\tan 60° = \dfrac{h+1.6}{d} = \dfrac{h+1.6}{h}$
$$\sqrt{3}h = h + 1.6$$$$h(\sqrt{3}-1) = 1.6$$$$h = \frac{1.6}{\sqrt{3}-1} = \frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{1.6(\sqrt{3}+1)}{2} = 0.8(\sqrt{3}+1)$$$$h = 0.8 \times 2.732 \approx 2.19 \text{ m}$$Or exactly: $h = 0.8(\sqrt{3}+1)$ m.
∴ Height of pedestal = $0.8(\sqrt{3}+1) \approx$ 2.19 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. #
Answer
Let h = height of building, d = distance between tower and building.
From foot of building: $\tan 60° = \dfrac{50}{d} \Rightarrow d = \dfrac{50}{\sqrt{3}}$
From foot of tower: $\tan 30° = \dfrac{h}{d}$
$$\frac{1}{\sqrt{3}} = \frac{h}{\frac{50}{\sqrt{3}}} = \frac{h\sqrt{3}}{50}$$$$h = \frac{50}{3} \approx 16.67 \text{ m}$$∴ Height of building = $\dfrac{50}{3} \approx$ 16.67 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 height of the poles and the distances of the point from the poles. #
Answer
Let h = height of each pole, and the point P be at distance x from one pole and (80-x) from the other.
$$\tan 60° = \frac{h}{x} \Rightarrow h = x\sqrt{3} \quad \cdots (1)$$$$\tan 30° = \frac{h}{80-x} \Rightarrow h = \frac{80-x}{\sqrt{3}} \quad \cdots (2)$$From (1) and (2):
$$x\sqrt{3} = \frac{80-x}{\sqrt{3}}$$$$3x = 80-x$$$$4x = 80 \Rightarrow x = 20 \text{ m}$$$$h = 20\sqrt{3} \text{ m}$$Distance from other pole = $80 - 20 = 60$ m.
∴ Height of poles = $20\sqrt{3} \approx$ 34.64 m; distances from poles = 20 m and 60 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. #
Answer
Let h = height, w = width of canal.
From directly opposite: $\tan 60° = \dfrac{h}{w} \Rightarrow h = w\sqrt{3} \quad \cdots (1)$
From 20 m farther: $\tan 30° = \dfrac{h}{w+20}$
$$\frac{1}{\sqrt{3}} = \frac{h}{w+20} \Rightarrow h = \frac{w+20}{\sqrt{3}} \quad \cdots (2)$$From (1) and (2):
$$w\sqrt{3} = \frac{w+20}{\sqrt{3}}$$$$3w = w+20$$$$2w = 20 \Rightarrow w = 10 \text{ m}$$$$h = 10\sqrt{3} \text{ m}$$∴ Height of tower = $10\sqrt{3} \approx$ 17.32 m; width of canal = 10 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. #
Answer
Let AB = building = 7 m, CD = cable tower, BC = horizontal distance.
Angle of depression of foot C = 45°: $\tan 45° = \dfrac{AB}{BC} \Rightarrow BC = 7$ m
Let DE = h (part of tower above building level), so CD = h + 7.
Angle of elevation to top D = 60°:
$$\tan 60° = \frac{DE}{BC} = \frac{h}{7}$$$$\sqrt{3} = \frac{h}{7} \Rightarrow h = 7\sqrt{3} \text{ m}$$Height of cable tower = $h + 7 = 7\sqrt{3} + 7 = 7(\sqrt{3}+1)$ m
∴ Height of cable tower = $7(\sqrt{3}+1) \approx$ 19.12 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. #
Answer
Height of lighthouse = 75 m. Let the two ships be at distances $d_1$ (nearer) and $d_2$ (farther).
Angle of depression 45°:
$$\tan 45° = \frac{75}{d_1} \Rightarrow d_1 = 75 \text{ m}$$Angle of depression 30°:
$$\tan 30° = \frac{75}{d_2} \Rightarrow d_2 = 75\sqrt{3} \text{ m}$$Distance between ships = $d_2 - d_1 = 75\sqrt{3} - 75 = 75(\sqrt{3}-1)$ m
∴ Distance between ships = $75(\sqrt{3}-1) \approx$ 54.9 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. #
Answer
Height of balloon above girl’s eyes = $88.2 - 1.2 = 87$ m.
At 60°: $\tan 60° = \dfrac{87}{d_1} \Rightarrow d_1 = \dfrac{87}{\sqrt{3}} = 29\sqrt{3}$ m
At 30°: $\tan 30° = \dfrac{87}{d_2} \Rightarrow d_2 = 87\sqrt{3}$ m
Distance travelled = $d_2 - d_1 = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}$ m
∴ Distance travelled = $58\sqrt{3} \approx$ 100.4 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. #
Answer
Let h = height of tower, and let the car’s position at first observation be A and at second observation be B.
$$\tan 30° = \frac{h}{d_1} \Rightarrow d_1 = h\sqrt{3}$$$$\tan 60° = \frac{h}{d_2} \Rightarrow d_2 = \frac{h}{\sqrt{3}}$$Distance covered in 6 seconds = $d_1 - d_2 = h\sqrt{3} - \dfrac{h}{\sqrt{3}} = h \cdot \dfrac{2}{\sqrt{3}} = \dfrac{2h}{\sqrt{3}}$
Speed = $\dfrac{2h}{6\sqrt{3}} = \dfrac{h}{3\sqrt{3}}$ m/s
Time to travel $d_2 = \dfrac{h}{\sqrt{3}}$:
$$t = \frac{d_2}{\text{speed}} = \frac{\frac{h}{\sqrt{3}}}{\frac{h}{3\sqrt{3}}} = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h} = 3 \text{ seconds}$$∴ Time taken = 3 seconds.