CBSE Class 10 Mathematics Application of Trignometry Worksheet Set A

SHORT ANSWER TYPE QUESTIONS :

Question. The angle of depression of the top and the bottom of a 9 m high building from the top of a tower are 30° and 60° respectively. Find the height of the tower and the distance between the building and the tower. 

Answer : Let AB represents the building and CD be the tower.
∴ AB = 9 m
In right Δ BDC, we have:
CD/DB = tan 60° = √3
⇒ CD = DB· √3 
In right Δ AEC, we have:
CE/AE = tan 30° = 1/√3
⇒ CD− 9/AE = 1/√3 ⇒ AE = √3 CD − 9√3
⇒ BD = √3 (DB·√3 ) − 9√3
⇒ BD = 3 BD − 9√3
⇒ 2BD = 9√3
⇒ BD = 9/2√3 = 9×1.732/2
⇒ BD = 7.8 m
From (1), we have,
CD = √3 × 9/2 × √3 = 27/2 = 13.5
Thus, height of the tower = 13.5 m
Distance between the building and the tower = 7.8 m

Question. A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower. 
Answer : In the figure, AB is the tower,
∴ AB = h metres
In rt Δ ABC, we have: 
BC/AC = tan 60°
⇒ h/20 = √3
|ä tan 60° = √3 and AB = 20 m
⇒ h = 20√3 metre
Thus, the height of the tower = 20√3 m.

Extra Based Questions :

Question. 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.
Answer : Let the tower be represented by AB in the figure.
Let AB = h metres.
∴ In right Δ ABC, we have:

Question. 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 height of the building be BC
∴ BC = 20 m
And height of the tower be CD.

⇒ 20+x/20 = √3 ⇒ 20 + x = 20√3
⇒ x = 20√3 − 20 = 20 [ √3 − 1]
⇒ x = 20 [1.732 − 1]
⇒ x = 20 × 0.732 = 14.64
Thus, the height of the tower is 14.64 m.

Question. 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 older 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 : In the figure, DE is the slide for younger children whereas AC is the slide for older children.
In right Δ ABC,
AB = 3 m
AC = length of the slide
∴ AB/AC = sin 60°
⇒ 3/AC = √3/2
⇒ AC = 2 × 3/√3 = 2√3 m

Again in right ΔBDE,
DE/BD = cosec 30° = 2
⇒ DE/1.5 = 2
⇒ DE = 2 × 1.5 m
⇒ DE = 3 m
Thus, the lengths of slides are 3 m and 2√3m.

Question. 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 : In the figure,

Question. 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 the original height of the tree = OP.
It is broken at A and its top is touching the ground at B.
Now, in right Δ AOB, we have

= 8/√3 + 16/√3 [∵ AB = AP]
= 24/√3m = 24/√3 × √3/√3 m = 8√3m

Question. 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 : In the figure, let AB is the height of the tower and C and D be the two positions of the car.

In right Δ ABD, we have:
AB/AD
= tan 60°
⇒ AB/AD = √3 ⇒ AB = √3 · AD
In right Δ ABC, we have:
AB/AC
= tan 30°
⇒ AB/AC = 1/√3 ⇒ AB = AC/√3
From (1) and (2)
√3 AD = AC/√3
⇒ AC = √3 × √3 × AD = 3 AD
Now CD = AC − AD
= 3 AD − AD = 2 AD
Since the distance 2 AD is covered in 6 seconds,
∴ The distance AD will be covered in 6/2 i.e., 3 seconds.
Thus, the time taken by the car to reach the tower from D is 3 seconds.

Question. 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 : Let in the right Δ AOB,

Question. 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 : In the figure, let AB represent the light house.
∴ AB = 75 m.
Let the two ships be C and D such that angles of depression from B are 45° and 30° respectively.
Now in right Δ ABC, we have:
AB/AC = tan 45°
⇒ 75/AC = 1 ⇒ AC = 75 
Again, in right Δ ABD, we have:
AB/AD = tan 30°
⇒ 75/AD = 1/√3 ⇒ AD = 75√3 
Since the distance between the two ships = CD
= AD − AC
= 75√3 − 75 = 75 [ √3 − 1]
= 75 [1.732 − 1] = 75 × 0.732 = 54.9
Thus, the required distance between the ships = 54.9 m.

Question. 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° (see figure).
Answer : In the figure, let AC is the rope and AB is the pole. In right Δ ABC, we have:

Thus, the required height of the pole is 10 m.

Question. 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 : In the figure, let height of the building = AB = h m
Let CD be the tower.

Question. 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 : In the figure, let AB be the height of the tower.

Question. The angles of elevation of the top of a tower from two points on the ground at distances a and b units from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is √ab units.
Answer : In the figure, AB is the tower, such that:

Question. The angles of depression of the top and the bottom of a building 50 m high as observed from the top of a tower are 30° and 60° respectively. Find the height of the tower and also the horizontal distance between the building and the tower.
Answer : In the figure,

Let AB = 50 m be the building.
Let CE be the tower such that CE = (50 + x) m
In right Δ ADE, we have:
DE/AD = tan 30° = 1/√3
⇒ x/AD = 1/√3 ⇒ AD ⇒ x √3 or BC = x √3
In right Δ ACE, we have:
CE/BC = tan 60° = 3√
⇒ 50 + x/BC = 3 ⇒ BC = 50 + x/3
From (1) and (2), we get
√3 x = 50 + x/3
⇒ √3x × √3 = 50 + x
⇒ 3x − x = 50 ⇒ x = 25
∴ Height of the tower = 50 + x
= 50 + 25
= 75 m
Now from (1), BC = √3 × x
= √3 × 25 m
= 1.732 × 25 m
= 43.25 m
i.e., The horizontal distance between the building and the tower = 43.25 m.

Question. From the top of a hill 200 m high, the angles of depression of the top and bottom of a pillar are 30° and 60° respectively. Find the height of the pillar and its distance from the hill. 
Answer : In the figure, let AD is the hill such that 

⇒ DE = √3×200/3 = 1.73×200/3
= 346/3 = 115.33 m
⇒ Distance between pillar and hill = 115.33 m
Now, BC = DE = 200/√3 m [∵ DE = BC]
In right Δ ABC, we have:
AB/BC = tan 30° = 1/√3
⇒ AB = BC/√3 = 200/√3 × 1/√3 = 200/3 [∵ BC = 200/√3]
= 66.67 m
∴ Height of the pillar
CE = AD − AB [∵ CE = BD]
= 200 − 66.67 m
= 133.33 m

Question. A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point ‘A’ on the ground is 60° and the angle of depression of the point ‘A’ from the top of the tower is 45°. Find the height of the tower. 
Answer : In the figure, let BC be the tower and CD be the pole.

Question. The length of the shadow of a tower standing on level plane is found to be 20 m longer when the sun’s altitude is 30° than when it was 60°. Find the height of the tower.
Answer : In the figure, let CD be the tower such that

Please click on below link to download CBSE Class 10 Mathematics Application of Trignometry Worksheet Set A