CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 14

Exercise-1 Subjective Questions

A-1. The angle of elevation of the top of a tower at a distance \( \dfrac{50\sqrt{3}}{3} \) metres from the foot is 60°. Find the height of the tower.
Answer:
\( h = \dfrac{50\sqrt{3}}{3}\tan 60^\circ = \dfrac{50\sqrt{3}}{3} \times \sqrt{3} = \) 50 m.
In simple words: Height = distance × tan 60°.

Teacher's Note:
1. tan 60° = √3.

A-2. If the altitude of the sun is 60°, then find the height of the vertical tower that will cast a shadow of length 20 m.
Answer:
\( h = 20\tan 60^\circ = \) \( 20\sqrt{3} \) m.
In simple words: Height = shadow × tan 60°.

Teacher's Note:
1. The altitude is the angle of elevation.

A-3. Find the length of the ladder making an angle of 45° with the ground and whose foot is 7 m away from the wall.
Answer:
\( \cos 45^\circ = \dfrac{7}{L} \Rightarrow L = 7\sqrt{2} = \) \( 7\sqrt{2} \) m.
In simple words: Foot distance over ladder length is cosine.

Teacher's Note:
1. cos 45° = 1/√2.

A-4. A man on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60°. Find the total time taken by the boat from the initial point to reach the shore.
Answer:
With cliff height h: initial distance \( h\sqrt{3} \), later distance \( \dfrac{h}{\sqrt{3}} \). In 6 min the boat covers \( h\sqrt{3} - \dfrac{h}{\sqrt{3}} = \dfrac{2h}{\sqrt{3}} \).
Remaining distance \( \dfrac{h}{\sqrt{3}} \) is half of that, so it takes 3 minutes. Total time \( = 6 + 3 = \) 9 minutes.
In simple words: The remaining distance is half of the distance covered.

Teacher's Note:
1. Speed is uniform, so time is proportional to distance.

A-5. The angles of elevation of the top of a tower from two points P and Q at distances of a and b respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is \( \sqrt{ab} \).
Answer:
\( \tan\theta = \dfrac{h}{a} \) and \( \tan(90^\circ - \theta) = \dfrac{h}{b} \). Multiplying gives \( 1 = \dfrac{h^2}{ab} \), so \( h = \sqrt{ab} \).
In simple words: Multiply the two tangents.

Teacher's Note:
1. The same result as the Board Level question.

A-6. Two pillars of equal height are on either side of a road which is 100 m wide. The angles of elevation of the top of the pillars are 60° and 30° at a point on the road between the pillars. Find the position of the point between the pillars. Also find the height of each pillar.
Answer:
Let the pillar height be h and the point be a m from the 60° pillar: \( a = \dfrac{h}{\sqrt{3}} \) and \( 100 - a = h\sqrt{3} \).
\( \dfrac{4h}{\sqrt{3}} = 100 \Rightarrow h = 25\sqrt{3} = \) 43.3 m. The point is \( a = 25 \) m from the first pillar and 75 m from the second.
In simple words: The nearer pillar has the larger angle.

Teacher's Note:
1. Add the two distances to get 100 m.

A-7. At a point, the angle of elevation of a tower is such that its tangent is \( \dfrac{5}{12} \). On walking 240 m nearer the tower, the tangent of the angle of elevation becomes \( \dfrac{3}{4} \). Find the height of the tower.
Answer:
\( h\cdot\dfrac{12}{5} - h\cdot\dfrac{4}{3} = 240 \Rightarrow h\cdot\dfrac{16}{15} = 240 \Rightarrow h = \) 225 m.
In simple words: Distance = h cot θ, and cot is the reciprocal of tan.

Teacher's Note:
1. The two distances differ by 240 m.

A-8. From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive kilometre stones on opposite sides of the aeroplane are observed to be \( \alpha \) and \( \beta \). Show that the height of the aeroplane above the road is \( \dfrac{\tan\alpha\tan\beta}{\tan\alpha + \tan\beta} \) kilometre.
Answer:
Distances from the foot of the aeroplane: \( h\cot\alpha \) and \( h\cot\beta \), which add up to 1 km.
\( h(\cot\alpha + \cot\beta) = 1 \Rightarrow h = \dfrac{1}{\cot\alpha + \cot\beta} = \dfrac{\tan\alpha\tan\beta}{\tan\alpha + \tan\beta} \).
In simple words: The two stones are 1 km apart on opposite sides.

Teacher's Note:
1. Multiply the top and bottom by tan α tan β.

A-9. The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/hr, find the constant height at which the jet is flying. [use \( \sqrt{3} = 1.732 \)]
Answer:
Speed \( = 720 \times \dfrac{5}{18} = 200 \) m/s, so the distance in 15 s is 3000 m.
\( h\cot 30^\circ - h\cot 60^\circ = 3000 \Rightarrow \dfrac{2h}{\sqrt{3}} = 3000 \Rightarrow h = 1500\sqrt{3} = \) 2598 m.
In simple words: Convert the speed to m/s first.

Teacher's Note:
1. The jet is flying away from A.

Exercise-1 Objective Questions

A-1. The angle of elevation of the top of a tower as observed from a point on the horizontal ground is x. If we move a distance d towards the foot of the tower, the angle of elevation increases to y, then the height of the tower is
(A) \( \dfrac{d\tan x\tan y}{\tan y - \tan x} \)   (B) \( d(\tan y + \tan x) \)   (C) \( d(\tan y - \tan x) \)   (D) \( \dfrac{d\tan x\tan y}{\tan y + \tan x} \)
Answer:
Ans. (A) \( \dfrac{d\tan x\tan y}{\tan y - \tan x} \)
\( h\cot x - h\cot y = d \Rightarrow h = \dfrac{d}{\cot x - \cot y} = \dfrac{d\tan x\tan y}{\tan y - \tan x} \).
In simple words: Multiply the top and bottom by tan x tan y.

Teacher's Note:
1. Distances are h cot x and h cot y.

A-2. The shadow of a tower, when the angle of elevation of the sun is 30°, is found to be 5 m longer than when it was 45°. The height of the tower in metres is
(A) \( \dfrac{5}{\sqrt{3} + 1} \)   (B) \( \dfrac{5}{2}(\sqrt{3} - 1) \)   (C) \( \dfrac{5}{2}(\sqrt{3} + 1) \)   (D) None of these
Answer:
Ans. (C) \( \dfrac{5}{2}(\sqrt{3} + 1) \)
\( h\cot 30^\circ - h\cot 45^\circ = 5 \Rightarrow h(\sqrt{3} - 1) = 5 \Rightarrow h = \dfrac{5}{\sqrt{3} - 1} = \dfrac{5}{2}(\sqrt{3} + 1) \).
In simple words: The shadow lengths are h cot 30° and h cot 45°.

Teacher's Note:
1. Rationalise 5/(√3 − 1).

A-3. The angles of elevation of the top of a vertical tower from two points 30 metres apart, and on the same straight line passing through the base of the tower, are 30° and 60° respectively. The height of the tower is
(A) 10 m   (B) 15 m   (C) \( 15\sqrt{3} \) m   (D) 30 m
Answer:
Ans. (C) \( 15\sqrt{3} \) m
\( h\cot 30^\circ - h\cot 60^\circ = 30 \Rightarrow \dfrac{2h}{\sqrt{3}} = 30 \Rightarrow h = 15\sqrt{3} \) m.
In simple words: Both distances are measured from the foot of the tower.

Teacher's Note:
1. Subtract the two distances.

A-4. A person on the top of a tower observes a scooter moving with uniform velocity towards the base of the tower. He finds that the angle of depression changes from 30° to 60° in 18 minutes. The scooter will reach the base of the tower in the next
(A) 9 minutes   (B) \( \dfrac{18}{\sqrt{3} - 1} \) minutes   (C) \( 6\sqrt{3} \) minutes   (D) the time depends upon the height of the tower
Answer:
Ans. (A) 9 minutes
The distance covered in 18 min is \( h\sqrt{3} - \dfrac{h}{\sqrt{3}} = \dfrac{2h}{\sqrt{3}} \); the remaining distance is \( \dfrac{h}{\sqrt{3}} \), half of it. Time \( = 9 \) minutes.
In simple words: Uniform speed means time is proportional to distance.

Teacher's Note:
1. Remaining distance is half the covered distance.

A-5. A ladder reaches a point on a wall which is 20 m above the ground and its foot is \( 20\sqrt{3} \) m away from the wall. The angle made by the ladder with the wall is
(A) 90°   (B) 60°   (C) 45°   (D) 30°
Answer:
Ans. (B) 60°
\( \tan\theta = \dfrac{20\sqrt{3}}{20} = \sqrt{3} \) where \( \theta \) is measured from the wall, so \( \theta = 60^\circ \).
In simple words: Foot distance is opposite the angle with the wall.

Teacher's Note:
1. The angle is measured from the wall.

Exercise-2 Objective Questions

Question 1. The upper part of a tree broken over by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree meets the ground is 15 m. The length of the broken part is
(A) 15 sin 30° m   (B) 15 cos 30° m   (C) 15 tan 30° m   (D) 15 sec 30° m
Answer:
Ans. (D) 15 sec 30° m
The broken part is the hypotenuse: \( \cos 30^\circ = \dfrac{15}{L} \Rightarrow L = 15\sec 30^\circ \).
In simple words: 1/cos is sec.

Teacher's Note:
1. Hypotenuse = adjacent ÷ cos.

Question 2. On level ground, the angle of elevation of the top of a tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is
(A) 10 m   (B) 15 m   (C) \( 10\sqrt{3} \) m   (D) 20 m
Answer:
Ans. (C) \( 10\sqrt{3} \) m
\( h\cot 30^\circ - h\cot 60^\circ = 20 \Rightarrow \dfrac{2h}{\sqrt{3}} = 20 \Rightarrow h = 10\sqrt{3} \) m.
In simple words: The two distances differ by 20 m.

Teacher's Note:
1. The same method as before.

Question 3. From a lighthouse 100 m high, it is observed that two ships are approaching it from west and south. If the angles of depression of the two ships are 30° and 45° respectively, the distance between the ships, in metres, is
(A) \( 100(\sqrt{3} + 1) \)   (B) \( 100(\sqrt{3} + 1)^2 \)   (C) 200   (D) 400
Answer:
Ans. (C) 200
The distances from the foot are \( 100\sqrt{3} \) and 100, along perpendicular directions. The distance between the ships is \( \sqrt{30000 + 10000} = 200 \) m.
In simple words: Use the Pythagoras theorem.

Teacher's Note:
1. West and south are at right angles.

Question 4. The length of a shadow of a pole is \( \sqrt{3} \) times the length of the pole. The angle of elevation of the sun is
(A) 30°   (B) 60°   (C) 90°   (D) 45°
Answer:
Ans. (A) 30°
\( \tan\theta = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ \).
In simple words: tan 30° = 1/√3.

Teacher's Note:
1. Height over shadow.

Question 5. If the angles of depression and elevation of the top of a tower of height h from the top and bottom of a second tower are x and y respectively, then the height of the second tower is
(A) \( h(\cot y + \cot x) \)   (B) \( h(\tan x + \tan y) \)   (C) \( h(1 + \tan x\cot y) \)   (D) \( h(\tan y\cot x + 1) \)
Answer:
Ans. (C) \( h(1 + \tan x\cot y) \)
The distance between the towers is \( d = h\cot y \). The second tower is taller than the first by \( d\tan x \), so its height is \( h + h\cot y\tan x = h(1 + \tan x\cot y) \).
In simple words: Read the question as: the first tower (height h) is seen from the second.

Teacher's Note:
1. Find the distance, then add the extra height.

Question 6. If the elevation of the sun changed from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high, made at these two positions, is
(A) 7.5 m   (B) 15 m   (C) \( 10\sqrt{3} \) m   (D) \( \dfrac{15}{\sqrt{3}} \) m
Answer:
Ans. (C) \( 10\sqrt{3} \) m
\( 15\cot 30^\circ - 15\cot 60^\circ = 15\sqrt{3} - 5\sqrt{3} = 10\sqrt{3} \) m.
In simple words: cot 30° = √3 and cot 60° = 1/√3.

Teacher's Note:
1. Shadow = height × cot.

Question 7. Vijay is walking towards a building. From his position the angles of elevation of the top of the electric rod on the top of the building and the top of the building are \( \phi \) and \( \theta \) respectively. If the heights of the electric rod and the building are p and q respectively, mark the correct statement.
(A) The height of the tower is \( \dfrac{p\tan\theta\tan\phi}{\tan\phi - \tan\theta} \)   (B) The height of the electric rod is \( \dfrac{q\tan\theta}{\tan\phi - \tan\theta} \)   (C) The height of the tower is \( \dfrac{p\tan\theta}{\tan\phi + \tan\theta} \)   (D) The height of the electric rod is \( \dfrac{q(\tan\phi - \tan\theta)}{\tan\theta} \)
Answer:
Ans. (D) The height of the electric rod is \( \dfrac{q(\tan\phi - \tan\theta)}{\tan\theta} \)
With distance d: \( q = d\tan\theta \) and \( p + q = d\tan\phi \). So \( p = d(\tan\phi - \tan\theta) = \dfrac{q}{\tan\theta}(\tan\phi - \tan\theta) \).
Only (D) is correct; (A) and (C) give wrong building heights and (B) has the tangent in the wrong place.
In simple words: Both angles are measured from the same point.

Teacher's Note:
1. Eliminate the distance d.

Question 8. The elevation of the top of a building under construction, at a point 120 m from the base, is 45°. How much higher should the building be raised so that the elevation becomes 60°?
(A) \( 120\sqrt{3} \) m   (B) \( 120(\sqrt{3} - 1) \) m   (C) \( 120(\sqrt{3} + 1) \)   (D) 100 m
Answer:
Ans. (B) \( 120(\sqrt{3} - 1) \) m
Present height 120 m; new height \( 120\sqrt{3} \). Increase \( = 120\sqrt{3} - 120 = 120(\sqrt{3} - 1) \) m.
In simple words: tan 45° = 1 and tan 60° = √3.

Teacher's Note:
1. Find the new height and subtract.

Question 9. There are 2 poles on a horizontal plane. From the mid-point of the line joining their feet, the tops of the poles appear at angles of elevation 60° and 30°. If the height of the first pole is 100 m, then the height of the second is
(A) \( \dfrac{100}{3} \) m   (B) \( \dfrac{100}{7} \) m   (C) \( 100\sqrt{3} \) m   (D) \( \dfrac{100}{\sqrt{3}} \) m
Answer:
Ans. (A) \( \dfrac{100}{3} \) m
Half the distance \( d = 100\cot 60^\circ = \dfrac{100}{\sqrt{3}} \). The second pole is \( d\tan 30^\circ = \dfrac{100}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} = \dfrac{100}{3} \) m.
In simple words: The 60° pole is the 100 m one.

Teacher's Note:
1. Both poles are the same distance from the mid-point.

Question 10. A person walking along a straight road observes that at the consecutive kilometre stones, the angles of elevation of a hill in front of him are 30° and 75°. Find the height of the hill. (Given \( \sin 75^\circ = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \))
(A) \( \dfrac{\sqrt{3} + 1}{4} \) km   (B) \( \dfrac{\sqrt{3} - 1}{4} \) km   (C) \( \dfrac{\sqrt{3} + 1}{2} \) km   (D) None of these
Answer:
Ans. (A) \( \dfrac{\sqrt{3} + 1}{4} \) km
In the triangle formed by the two stones A, B and the top P: \( \angle PBA \) is the supplement of 75° and \( \angle APB = 75^\circ - 30^\circ = 45^\circ \). By the sine rule \( \dfrac{BP}{\sin 30^\circ} = \dfrac{1}{\sin 45^\circ} \Rightarrow BP = \dfrac{1}{\sqrt{2}} \).
Height \( = BP\sin 75^\circ = \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3} + 1}{2\sqrt{2}} = \dfrac{\sqrt{3} + 1}{4} \) km.
In simple words: An exterior angle is the sum of the opposite interior angles.

Teacher's Note:
1. The angle at the hill is 75° − 30° = 45°.