CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 10

Very Short Answer Type Questions

Question 1. As shown in the figure, PQ is a 6 m high pole and CD is a ladder inclined at an angle of \( 60^\circ \) to the horizontal and reaches up to a point D of the pole. If \( PD = 2.54 \) m, find the length of the ladder. (use \( \sqrt{3} = 1.73 \))

Answer:
\( QD = PQ - PD = 6 - 2.54 = 3.46 \) m.
In right \( \triangle DQC \): \( \sin 60^\circ = \frac{QD}{CD} \Rightarrow \frac{1.73}{2} = \frac{3.46}{CD} \Rightarrow CD = \frac{3.46 \times 2}{1.73} = 4 \) m.
The length of the ladder is 4 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 2. A ladder, leaning against a wall, makes an angle of \( 60^\circ \) with the ground. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.
Answer:
Let the ladder be L m long. The foot is 2.5 m from the wall, which is the side adjacent to the \( 60^\circ \) angle.
\( \cos 60^\circ = \frac{2.5}{L} \Rightarrow \frac{1}{2} = \frac{2.5}{L} \Rightarrow L = 5 \).
The length of the ladder is 5 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 3. The ratio of the length of a vertical rod and the length of its shadow is \( 1 : \sqrt{3} \). Find the angle of elevation of the sun at that moment. [CBSE Delhi Set-I, 2020]
Answer:
\( \tan \theta = \frac{\text{rod}}{\text{shadow}} = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ \).
The angle of elevation of the sun is \( 30^\circ \).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 4. A tower PQ is 20 m high and QR, its shadow on the ground, is \( 20\sqrt{3} \) m long. Find the angle of elevation of the sun's altitude.
Answer:
\( \tan \theta = \frac{PQ}{QR} = \frac{20}{20\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ \).
The angle of elevation of the sun is \( 30^\circ \).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 5. A pole casts a shadow of length \( 2\sqrt{3} \) m on the ground, when the sun's elevation is \( 60^\circ \). Find the height of the pole.
Answer:
\( \tan 60^\circ = \frac{h}{2\sqrt{3}} \Rightarrow \sqrt{3} = \frac{h}{2\sqrt{3}} \Rightarrow h = 6 \).
The height of the pole is 6 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 6. An observer, 1.7 m tall, is \( 20\sqrt{3} \) m away from a building. The angle of elevation from the eye of the observer to the top of the building is \( 30^\circ \). Find the height of the building.
Answer:
The height of the building above the observer's eye is \( 20\sqrt{3} \times \tan 30^\circ = 20\sqrt{3} \times \frac{1}{\sqrt{3}} = 20 \) m.
Height of the building \( = 20 + 1.7 = 21.7 \) m.
The height of the building is 21.7 m.

Teacher's Note:
1. The angle is measured from the eye of the observer, so add the observer's height at the end.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 7. The shadow of a building is 20 m long when the angle of elevation of the sun is \( 60^\circ \). Find the height of the building.
Answer:
\( \tan 60^\circ = \frac{h}{20} \Rightarrow h = 20\sqrt{3} \).
The height of the building is \( 20\sqrt{3} \) m (about 34.64 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 8. A tower is \( 50\sqrt{3} \) m high. Find the angle of elevation, if its top from a point 50 m away from its foot.
Answer:
\( \tan \theta = \frac{50\sqrt{3}}{50} = \sqrt{3} \Rightarrow \theta = 60^\circ \).
The angle of elevation is \( 60^\circ \).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 9. A balloon is connected to a meteorological station by a cable of length 200 m, inclined at \( 60^\circ \) to the horizontal. Find the height of the balloon from the ground. Assume that there is no slack in the cable.
Answer:
\( \sin 60^\circ = \frac{h}{200} \Rightarrow h = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \).
The height of the balloon is \( 100\sqrt{3} \) m (about 173.2 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Short Answer Type Questions

Question 10. The tops of two towers of height x and y units, standing on the ground, subtend angles of \( 30^\circ \) and \( 60^\circ \) respectively at the centre of the line joining their feet. Find \( x : y \). [CBSE Delhi Set-I, II, III, 2015]

Answer:
Let the centre of the line joining the feet be at distance d from each tower.
\( \tan 30^\circ = \frac{x}{d} \Rightarrow x = \frac{d}{\sqrt{3}} \) and \( \tan 60^\circ = \frac{y}{d} \Rightarrow y = \sqrt{3}d \).
\( x : y = \frac{d}{\sqrt{3}} : \sqrt{3}d = 1 : 3 \).
Hence \( x : y = 1 : 3 \).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 11. If the angle of elevation of a cloud from a point 200 m above a pond is \( 30^\circ \) and the angle of depression of its reflection in the pond is \( 60^\circ \), then find the height of the cloud above the pond.
Answer:
Let the cloud be h m above the pond; its reflection is h m below the pond surface. The observer is 200 m above the pond, so the cloud is \( (h - 200) \) m above the observer and its reflection is \( (h + 200) \) m below the observer. Let the horizontal distance be d.
\( \tan 30^\circ = \frac{h - 200}{d} \Rightarrow d = \sqrt{3}(h - 200) \) and \( \tan 60^\circ = \frac{h + 200}{d} \Rightarrow d = \frac{h + 200}{\sqrt{3}} \).
\( \sqrt{3}(h - 200) = \frac{h + 200}{\sqrt{3}} \Rightarrow 3h - 600 = h + 200 \Rightarrow h = 400 \).
The height of the cloud above the pond is 400 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 12. A man on a hill observes a fishing trawler at an angle of depression of \( 30^\circ \) which is moving towards the shore to the point immediately beneath the observer with a uniform speed. 6 minutes later, the angle of depression of the trawler is found to be \( 60^\circ \). Find the time taken by the trawler to reach the shore.
Answer:
Let H be the height of the observer above the sea. The horizontal distance of the trawler is \( H\cot 30^\circ = H\sqrt{3} \) at first and \( H\cot 60^\circ = \frac{H}{\sqrt{3}} \) later.
Distance covered in 6 minutes \( = H\sqrt{3} - \frac{H}{\sqrt{3}} = \frac{2H}{\sqrt{3}} \).
Remaining distance \( = \frac{H}{\sqrt{3}} \), which is half of \( \frac{2H}{\sqrt{3}} \). At a uniform speed it takes half of 6 minutes.
The trawler takes 3 minutes more to reach the shore.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 13. A boat is being rowed away from a hill 150 m high. Find the speed of the boat, if the angle of depression of the boat from the top of the hill changes from \( 60^\circ \) to \( 45^\circ \) in 2 minutes.
Answer:
Horizontal distance of the boat at \( 60^\circ \): \( 150\cot 60^\circ = \frac{150}{\sqrt{3}} = 50\sqrt{3} \) m. At \( 45^\circ \): \( 150\cot 45^\circ = 150 \) m.
Distance covered in 2 minutes \( = 150 - 50\sqrt{3} = 50(3 - \sqrt{3}) \approx 63.4 \) m.
Speed \( = \frac{50(3 - \sqrt{3})}{2} = 25(3 - \sqrt{3}) \) m/min, about 31.7 m/min (about 1.9 km/h).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 14. From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as \( 60^\circ \) and \( 45^\circ \). Find the distance between the two cars. [Delhi/OD Compt. Set-I, II, III, 2017]

Answer:
Let the cars be at distances \( d_1 \) and \( d_2 \) from the foot of the tower. \( d_1 = 120\cot 60^\circ = \frac{120}{\sqrt{3}} = 40\sqrt{3} \) m and \( d_2 = 120\cot 45^\circ = 120 \) m.
Since the cars are on opposite sides of the tower, the distance between them is \( d_1 + d_2 = 120 + 40\sqrt{3} = 40(3 + \sqrt{3}) \) m.
The distance between the cars is \( 40(3 + \sqrt{3}) \) m (about 189.3 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.