Class 10 Maths Chapter 9 Applications of Trigonometry Objective Questions

 

Applications of Trigonometry

1. 1. A technician has to repair a light on a pole of a height of 10 m. She needs to reach a point 1 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use when inclined at an angle of 60∘ to the ground that would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder?

Answer: (D) 6√3 m

Solution: The given situation is explained in the figure below:

DB – DC = CB

⇒BC=9msin60°=BCAC⇒AC=BCsin60°=

∴ The height of the ladder should be 6√3 m.

2. 2. A statue, 2 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: (B) 2(√3 – 1)

Solution:

CD = 2m

Let BC = x

AB = x (using tan 45∘)

In ΔABD,

BD = AB √3 = x √3

We know that CD = BD – BC = 2m

X(√3 – 1)

X = 2(√3 -1 )

Therefore, the height of the pedestal is 2(√3 -1 ).

3. 3. 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 60 m high, find the height of the building.
      1. 30 m
      2. 40 m
      3. 20 m
      4. 10 m

Answer: (C) 20 m

Solution:

The given situation is explained in the figure above.

tan60°=DC/BC⇒BC=DC/tan60°= 60/ √3 = 20 √3 m

tan30°=AB/BC

⇒AB=BC×tan30°=20 √3 × (1/√3 m) = 20 m

Thus, the height of the building is 20 m.

4. 4. A TV tower stands vertically on a bank of a canal, with a height of 10 √3 m. 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 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 distance between the opposite bank of the canal at the point with a 30° angle of elevation.
      1. 30 m
      2. 20 m
      3. 45 m
      4. 35 m

Answer: (B) 20 m

Solution:

The above figure represents the situation given in the question.

tan60°=AB/ BC

⇒AB=BCtan60°= BC √3……….(1)

⇒BC=AB/tan60°=AB/ tan60°

tan30°=AB/BD=AB/ (CD+BC)

⇒DC+BC=AB/tan30°=AB √3

⇒DC=AB

⇒DC = 20 m

Hence, the distance between the opposite bank of the canal is 20 m.

5. 5. As observed from the top of a 150 m high lighthouse from the sea level, the angles of depression of the 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: (B) 150 (√3 – 1)

Solution:

Here, the lighthouse, BC = 150 m

In ΔBDC,

BD = BC = 150 m (using tan 45°)

In ΔABC,

AB = BC √3 (using tan 30°)

AB= 150√3

So, distance AD = 150 (√3 – 1)

Hence, the between the two ships is 150 (√3 – 1).

6. 6. An observer of √3m tall is 3 m away from a pole of 2√3 high. What is the angle of elevation of the top of the pole from the observer?
      1. 60°
      2. 30°
      3. 45°
      4. 90°

Answer: (B) 30°

Solution:

Height of the pole that is above man’s height = 2 √3 – √3 = √3 m

Hence, AB = √3 m

BC = 3 m

Hence, tan C = AB/BC =

⟹ C, angle of elevation = 30°

Therefore, the angle of elevation of the top of the pole from the observer is 30°.

Introduction

7. 7. 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.

Solution: In ΔABC, taking the tangent of ∠C, we have,

tan C=AB/BC

⇒tan 30°=h/30

⇒

$\begin{array}{l}\frac{1}{\sqrt{3}}\end{array}$

= h/30

=10(1.732)

=17.32m = 10√3 m

Hence, the height of the tower is 10√3 metres.

8. 8. An observer having 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from her eyes is 45°. What is the height of the tower?
      1. 20 m
      2. 10 m
      3. 40 m
      4. 30 m

Answer: (D) 30 m

Solution: Let AB be the tower of height h, and CD be the observer of height, 1.5 m at a distance of 28.5 m from tower AB.

In ΔAED, we have

tan45°=h/28.5

⇒1=h/28.5

⇒h=28.5 m

∴ h=AE+BE=AE+DC

= (28.5+1.5) m=30 m

Height of tower = h + 1.5

= 28.5 + 1.5

= 30 m

Hence, the height of the tower is 30 m.

9. 9. An electrician has to repair an electrical fault on a pole of a height of 4 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use when inclined at an angle of 60° to the horizontal that would enable him to reach the required position?

Answer: (A) 

Solution: Let AC be the electric pole of height 4 m. Let B be a point 1.3 m below the top A of the pole AC.

Then, BC = AC – AB = (4 – 1.3) m = 2.7 m

Let BD be the ladder inclined at an angle of 60° to the horizontal.

In ΔBCD, we have

sin60°=2.7/L.

Hence, the length of the ladder should be m.

10. 10. 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°.
        1. 10 m
        2. 15 m
        3. 20 m
        4. 35 m

Answer: (A) 10 m

Solution: Let AB be the vertical pole and CA be the 20 m long rope, such that one end is tied from the top of the vertical pole AB and the other end C is tied to a point C on the ground.

In ΔABC, we have

sin 30°=h/20

⇒1/2=h/20

⇒h=10 m

Hence, the height of the pole is 10 m.

Heights and Distances

11. 11. An observer of 2.25 m tall is 42.75 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?
        1. 40 m
        2. 50 m
        3. 45 m
        4. 35 m

Answer: (C) 45 m

Solution:

The given situation is explained in the figure above:

In triangle ABE,

tan45°=AB/EB

Also, EB=DC

∴ tan45°=AB/ DC

⇒AB=DC × tan 45°

⇒AB=1×42.75

Hence, the height of the chimney = AC = AB + BC

We can observe that BC = ED.

Thus, AC = AB + ED

= 42.75 + 2.25

= 45 m.

Hence, the height of the chimney is 45 m.

12. 12. A tower stands vertically on the ground. From a point on the ground, which is 30 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 30°. Find the height of the tower.

Answer: (B) 10√3 m

Solution: The given situation is explained in the Δ below:

Now, tan30°=AB/ BC

∴ The height of the tower is  10√3 m.

13. 13. The angles of depression of the top and the bottom of a 10 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building.

Solution:

The above figure represents the situation aptly.

(ED is the height of the building)

 

14. 14. 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 the height of 2 m and is inclined at an angle of 30° to the ground. What should be the length of the slide?
        1. 4
        2. 2
        3. 1.5
        4. 3

Answer: (A) 4

Solution: The given situation can be explained using the figure below:

In the right-angled triangle ABC,

sin∠ABC=AC/AB=1/2

⇒sin30°=2/AB⇒AB= 2/ (1/2)

⇒AB = 4m

∴ The length of the slide is 4m.

15. 15. A kite is flying at a height of 30 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.

Solution: The situation can be explained using the figure below:

In the given right-angled triangle:

sin (∠ACB) =AB/AC

⇒sin60°=AB/AC

⇒AC=AB/sin60°=

∴ The length of the string is 20√3 m

16. 16. A vertical pole of 30 m is fixed on a tower. From a point on the level ground, the angles of elevation of the top and bottom of the pole are 60° and 45°. Find the height of the tower.

Answer: (B) 15 (√3 + 1)

Solution:

The situation can be explained by the figure above.

Therefore, the height of the tower is 15 (√3 + 1).

17. 17. The value of tan A +sin A=M, and tan A – sin A=N.

The value of (M2−N2) / (MN) 0.5

17. 17. 1. 4
        2. 3
        3. 2
        4. 1

Answer: (A) 4

Solution: M2-N2 = (Tan A+ Sin A + Tan A –Sin A) (Tan A +Sin A – Tan A+ Sin A)

M2-N2 =4 tan A sin A

And (MN) 0.5 = (tan2A−sin2A) 0.5

18. 18. Two towers, A and B, are standing at some distance apart. From the top of tower A, the angle of depression of the foot of tower B is found to be 30°. From the top of tower B, the angle of depression of the foot of tower A is found to be 60°. If the height of tower B is ‘h’ m, then the height of tower A in terms of ‘h’ is _____ m

Answer: (B) h/3 m

Solution:

Let the height of tower A be = AB = H.

And the height of tower B = CD = h

In triangle ABC,

tan30° = AB/AC = H/AC ……………………………. 1

In triangle ADC,

tan60° = CD/AC = h/AC………………………………….2

Divide 1 by 2

We get tan30°/tan60° = H/h

Therefore, H = h/3 m.

19. 19. A 1.5 m tall boy is standing some distance from a 31.5 m tall building. If he walks ’d’ m towards the building, the angle of elevation of the top of the building changes from 30° to 60°. Find the length of d. (Take √3 = 1.73)
        1. 30.15 m
        2. 38.33 m
        3. 22.91 m
        4. 34.55 m

Answer: (D) 34.55 m

Solution:

The above figure explains the situation given in the question.

∴ The distance moved by the boy, d, is 34.55 m.

20. 20. The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be 45° and 30°. Find the distance between the two objects. (Take √3 = 1.73,)
        1. 200 m
        2. 150 m
        3. 107.5 m
        4. 73.2 m

Answer: (D) 73.2 m

Solution: Let C and D be the objects and CD be the distance between the objects.

In ΔABC, tan 45° = AB/AC

AB=AC=100 m

In ΔABD, tan 30° = AB/AD

CD=AD−AC=173.2−100=73.2 metres

Hence, the distance between the two objects is 73.2 metres.