Question

the two boys are on opposite sides of tower of 100 metres height. they measure the angle of elevation of the top of the tower as 30 degrees &45 degrees respectively . find the distance through which the boys are separated ​

Answer 1

Answer:

tan theta is equals to 10 by 10 root 3 is equals to 1 by root 3 is equals to theta is equal to 30 degree

Answer 2

Therefore the distance between the boy-1 and boy-2 is 100(√3 + 1 ) m

Given:

Height of the tower = 100 m

The angle of elevation of the tower concerning boy-1 = 30°

The angle of elevation of the tower concerning boy-2 = 45°

To Find:

The distance between the boys.

Solution:

The given question can be solved very easily as shown below.

Given that,

Height of the tower = h = 100 m

The angle of elevation of the tower concerning boy-1 = θ₁ = 30°

The angle of elevation of the tower concerning boy-2 = θ₂ = 45°

Let the distance between the boy-1 and the tower = x₁

Let the distance between the boy-2 and the tower = x₂

The distance between the boy-1 and the tower is given by,

⇒ Tan θ₁ = Opposite/Adjacent = h/x₁

⇒ Tan 30° = 100/ x₁

⇒ 1/√3 = 100/x₁ ⇒  x₁ = 100√3 m

Now,  the distance between the boy-2 and the tower is given by,

⇒ Tan θ₂ = h/x₂

⇒ Tan 45° = 100/x₂

⇒ 1 = 100/x₂ ⇒ x₂ = 100 m

The distance between the boy-1 and boy-2 = x₁ + x₂ = 100√3 + 100 = 100 ( √3 + 1 ) m

Therefore the distance between the boy-1 and boy-2 is 100(√3 + 1 ) m

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