#### Question

A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are 30° and 60° respectively. What is the distance the two cars and how far is each car from the tower?

#### Solution

Let AB be the height of tower 50 m and angle of depression from the top of the tower are 60° and 30° respectively at two observing Car C and D.

Let BD = x m, CD = y m and ∠ADB = 30°, ∠ACb = 60°

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle ABD

`=> tan D = (AB)/(BD)`

`=> tan 30° = 50/x`

`=> 1/sqrt3 = 50/x`

`=> x = 50sqrt3`

Since x = 86.6

Again in a triangle ABC

`=> tan C = (AB)/(BC)`

`=> tan 60° = 50/(x - y)`

`=> sqrt3 = 50/(x - y)`

`=> sqrtt3 xx 50sqrt3 - sqrt3y = 50`

`=> y = 57.67`

Therefore x - y= 86.6 - 57.67

`=> x - y = 28.93`

Hence the distance of first car from tower is 86.6 m

And the distance of second car from tower is 57.67 m

And the distance between cars is 28.93 m