Class 10 Maths Chapter 9 Important Question Answers - Some Application of Trigonometry

Very Short Answer Type Questions

Q5: In the given figure, AB is a 6 m high pole and DC is a ladder inclined at an angle of 60° to the horizontal and reaches up to point D of pole. If AD = 2.54 m, find the length of ladder. (use √3 = 1.73)

Ans: 

sin 60° = BDDC

√32 = 3.46DC

DC = 3.46 × 2√3 = 3.461.73 = 4

Thus length of ladder is 4 m.

Q6: An observer 1.5 m tall is 28.5 m away from a tower 30 m high. Find the angle of elevation of the top of the tower from his eye.

Ans: As per given in question we have drawn figure below.

Here,

AE = 1.5 m is height of observer and BD = 30 m is tower.

Now,

BC = 30 - 1.5 = 28.5 m

In △BAC,

tan θ = BCAC

tan θ = 28.528.5 = 1 = tan 45°

θ = 45°

Hence, angle of elevation is 45°.

Q7: If a tower 30 m high, casts a shadow 10√3 m long on the ground, then what is the angle of elevation of the sun?
Ans: Let required angle be θ.

tan θ = 3010 √3

tan θ = √3

⇒ tan θ = tan 60°

∴ θ = 60°

Q8: The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x : y.
Ans: When base is same for both towers and their heights are given, i.e., x and y respectively

Let the base of towers be k.

tan 30° = xk     |     tan 60° = yk

x = k tan 30° = k√3     ...(i)

y = k tan 60° = k √3     ...(ii)

From equations (i) and (ii),

xy = k√3k √3 = kk × 1√3 = 13 = 1 : 3

Q9: A pole 6 m high casts a shadow 2√3 m long on the ground, then find the Sun’s elevation.
Ans: Let the Sun’s elevation be θ. As per given in question we have drawn figure below.

Length of pole is 6 m and length of shadow is 2√3 m.

From △ABC we have,

tan θ = ABBC = 62 √3 = 3√3 = √3 = tan 60°

θ = 60°

Hence sun’s elevation is 60°.