Question

An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. [Use `sqrt(3) = 1.732`.]

Answer 1

Let CD be the height of the aeroplane above the river at some instant. Suppose A and B be two points on both banks of the river in opposite directions

Height of the aeroplane above the river, CD = 300 m
Now,

∠CAD = ∠ADX = 60º (Alternate angles)

∠CBD = ∠BDY = 45º (Alternate angles)

In right ∆ACD,

`tan 60^@ = (CD)/(AC)`

`=> sqrt3 = 300/(AC)`

`=> AC = 300/sqrt3 = 100sqrt3 m`

In right ∆BCD

`tan 45^@ = (CD)/(BC)`

`=> 1 = 300/(BC)`

=> BC = 300 m

∴ Width of the river, AB

= BC + AC

`= 300 + 100sqrt3`

`= 300 + 100 xx 1.73`

= 300 + 173

= 473 m

Thus, the width of the river is 473 m