Question

An aeroplane is flying at a height of 210 m. Flying at this height at some instant the angles of depression of two points in a line in opposite directions on both the banks of the river are 45° and 60°. Find the width of the river. (Use `sqrt(3) = 1.73`)

Answer 1

Given: An aeroplane is at height PM = 210 m. The angles of depression to two points A and B on opposite banks are 60° and 45° respectively (points A and B lie on a line through M, the foot of the plane).

Step-wise calculation:

1. Let AM = x (distance from M to the point with 60° depression) and BM = y (distance from M to the point with 45° depression).

Then width of the river AB = x + y.

2. From triangle PMA (angle of depression 60°):

`tan 60° = "PM"/"AM"` 

⇒ `sqrt(3) = 210/x` 

⇒ `x = 210/sqrt(3)`

= `210/1.73`   ...(Using `sqrt(3) = 1.73`)

= 121.39 m 

3. From triangle PMB (angle of depression 45°):

`tan 45^circ = "PM"/"BM"` 

⇒ `1 = 210/y` 

⇒ y = 210 m

4. Width AB = x + y

= 121.39 + 210

= 331.39 m ≈ 331.4 m.

The width of the river is approximately 331.4 metres.