CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 12

Type I: Short Answer Type Questions (2 Marks Each)

Question 1. A pole 6 m high casts a shadow \( 2\sqrt{3} \) m long on the ground. Find the Sun's elevation.
Answer:
\( \tan\theta = \dfrac{6}{2\sqrt{3}} = \sqrt{3} \Rightarrow \theta = \) 60°.
In simple words: Height divided by shadow gives the tangent.

Teacher's Note:
1. tan 60° = √3.

Question 2. Find the angle of elevation of the sun when the shadow of a pole is equal to the height of the pole.
Answer:
\( \tan\theta = \dfrac{h}{h} = 1 \Rightarrow \theta = \) 45°.
In simple words: Equal height and shadow mean 45°.

Teacher's Note:
1. tan 45° = 1.

Question 3. A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.
Answer:
The angle with the wall is 60°, so the height is the side adjacent to it: \( h = 15\cos 60^\circ = \) 7.5 m.
In simple words: The angle is with the wall, not the ground.

Teacher's Note:
1. The height is next to the angle, so use cosine.

Question 4. An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Answer:
The height of the tower above the eye is \( 22 - 1.5 = 20.5 \) m, so \( \tan\theta = \dfrac{20.5}{20.5} = 1 \Rightarrow \theta = \) 45°.
In simple words: Subtract the observer's height first.

Teacher's Note:
1. The observer's eye is 1.5 m above the ground.

Question 5. The angle of elevation of the top of a tower from two points at distances s and t from its foot are complementary. Prove that the height of the tower is \( \sqrt{st} \).
Answer:
Let the angles be \( \theta \) and \( 90^\circ - \theta \), and the height h. \( \tan\theta = \dfrac{h}{s} \) and \( \tan(90^\circ - \theta) = \cot\theta = \dfrac{h}{t} \).
Multiplying: \( \tan\theta\cot\theta = 1 = \dfrac{h^2}{st} \Rightarrow h = \sqrt{st} \).
In simple words: Multiply the two tangents.

Teacher's Note:
1. tan θ × cot θ = 1.
2. The two points are on the same side of the tower.

Question 6. From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45° and 60°. If the cars are 100 m apart, find the height of the balloon.
Answer:
Take the cars on the same side. With height h: \( h\cot 45^\circ - h\cot 60^\circ = 100 \Rightarrow h\left(1 - \dfrac{1}{\sqrt{3}}\right) = 100 \).
\( h = \dfrac{100\sqrt{3}}{\sqrt{3} - 1} = 50\sqrt{3}(\sqrt{3} + 1) = \) \( 50(3 + \sqrt{3}) \) m (about 236.6 m).
In simple words: The nearer car is seen at the larger angle.

Teacher's Note:
1. The cars are on the same side of the balloon.
2. Rationalise the denominator.

Question 7. The shadow of a tower standing on a level plane is found to be 50 m longer when the Sun's elevation is 30° than when it is 60°. Find the height of the tower.
Answer:
\( h\cot 30^\circ - h\cot 60^\circ = 50 \Rightarrow h\left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = 50 \Rightarrow \dfrac{2h}{\sqrt{3}} = 50 \).
\( h = \) \( 25\sqrt{3} \) m.
In simple words: Shadow = h cot θ.

Teacher's Note:
1. A lower sun gives a longer shadow.

Type II: Long Answer Type Questions (3 Marks Each)

Question 8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are \( \alpha \) and \( \beta \) respectively. Prove that the height of the tower is \( \dfrac{h\tan\alpha}{\tan\beta - \tan\alpha} \).
Answer:
Let the tower be H and the point be at distance x. \( \tan\alpha = \dfrac{H}{x} \) and \( \tan\beta = \dfrac{H + h}{x} \).
So \( x = \dfrac{H}{\tan\alpha} = \dfrac{H + h}{\tan\beta} \Rightarrow H\tan\beta = (H + h)\tan\alpha \Rightarrow H(\tan\beta - \tan\alpha) = h\tan\alpha \).
\( H = \dfrac{h\tan\alpha}{\tan\beta - \tan\alpha} \).
In simple words: Equate the two expressions for the distance.

Teacher's Note:
1. The distance x is common to both right triangles.

Question 9. The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
Answer:
The new angle is 45°. \( h\cot 30^\circ - h\cot 45^\circ = 20 \Rightarrow h(\sqrt{3} - 1) = 20 \).
\( h = \dfrac{20}{\sqrt{3} - 1} = \) \( 10(\sqrt{3} + 1) \) m.
In simple words: 30° + 15° = 45°.

Teacher's Note:
1. The two distances differ by 20 m.

Question 10. From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower, are \( \alpha \) and \( \beta \) (\( \beta > \alpha \)). Find the distance between the two objects.
Answer:
The nearer object is seen at the larger angle \( \beta \). Distances from the foot: \( h\cot\beta \) and \( h\cot\alpha \).
Distance between them \( = \) \( h(\cot\alpha - \cot\beta) \).
In simple words: The farther object has the smaller depression angle.

Teacher's Note:
1. Both objects are on the same side of the tower.

Question 11. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.
Answer:
Let the horizontal distance be d and the tower height H. \( H = d\sqrt{3} \) and \( H - 10 = d \).
\( d(\sqrt{3} - 1) = 10 \Rightarrow d = 5(\sqrt{3} + 1) \), so \( H = \sqrt{3}d = \) \( 5(3 + \sqrt{3}) \) m.
In simple words: Both points are the same distance from the tower.

Teacher's Note:
1. The second point is higher by 10 m, so it is 10 m closer to the top.

Question 12. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At a certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon above the ground.
Answer:
Let the balloon be at height H and horizontal distance d from the house. The windows are at 2 m and 6 m: \( H - 2 = d\sqrt{3} \) and \( H - 6 = \dfrac{d}{\sqrt{3}} \).
Subtracting: \( 4 = d\left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = \dfrac{2d}{\sqrt{3}} \Rightarrow d = 2\sqrt{3} \). \( H = 2 + 2\sqrt{3} \times \sqrt{3} = \) 8 m.
In simple words: The upper window is 6 m above the ground.

Teacher's Note:
1. 4 m above the lower window means 2 + 4 = 6 m.

Question 13. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.
Answer:
Distance \( d = \dfrac{30}{\tan 60^\circ} = \dfrac{30}{\sqrt{3}} = \) \( 10\sqrt{3} \) m.
Height of the other tower \( = d\tan 30^\circ = 10\sqrt{3} \times \dfrac{1}{\sqrt{3}} = \) 10 m.
In simple words: The two towers give two right triangles with the same base.

Teacher's Note:
1. Use the first tower to get the distance.

Type III: Very Long Answer Type Question (4 Marks)

Question 14. A ladder rests against a vertical wall at an inclination \( \alpha \) to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle \( \beta \) with the horizontal. Show that \( \dfrac{p}{q} = \dfrac{\cos\beta - \cos\alpha}{\sin\alpha - \sin\beta} \).
Answer:
Let the ladder length be L. Initially the foot is at \( L\cos\alpha \) and the top at \( L\sin\alpha \). Finally the foot is at \( L\cos\beta \) and the top at \( L\sin\beta \).
\( p = L\cos\beta - L\cos\alpha = L(\cos\beta - \cos\alpha) \) and \( q = L\sin\alpha - L\sin\beta = L(\sin\alpha - \sin\beta) \).
Dividing: \( \dfrac{p}{q} = \dfrac{\cos\beta - \cos\alpha}{\sin\alpha - \sin\beta} \).
In simple words: The ladder length stays the same.

Teacher's Note:
1. Write the foot distance and the top height before and after.