Question

From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as 60° and 45°. Find the distance between the cars. (Take `sqrt(3) = 1.732`)

Answer 1

Given: From the top of a 120 m tower the angles of depression to two cars on opposite sides, collinear with the base are 60° and 45°.

Step-wise calculation:

1. Let the horizontal distances of the cars from the base be d₁ (for 60°) and d₂ (for 45°).

2. Use tan θ = `"Opposite"/"Adjacent"` (angle of depression = angle of elevation from ground):

For 60°: `tan 60^circ = sqrt(3) = 1.732` 

⇒ d₁ = `"Height"/(tan 60)` 

= `120/1.732` 

= `120/sqrt(3)` 

= `40sqrt(3)` 

Using `sqrt(3) = 1.732` gives d₁ = 40 × 1.732 = 69.28 m.

For 45°: tan 45° = 1

⇒ d₂ = `120/1`

= 120 m

3. Cars are on opposite sides.

So distance between them = d₁ + d₂

= 69.28 + 120

= 189.28 m

The distance between the two cars is 189.28 m.