Question

A man watches his friend swimming in his apartment swimming pool from the roof of the building which is 50 m high. When the swimmer was at one end, the angle of depression was ${40}^{\circ }$. When the swimmer reaches the other end, the angle of depression changes to 65${}^{\circ }$. The distance covered by the swimmer is

$[Tan{65}^{\circ }=2.14,Tan{40}^{\circ }=0.83]$

• A. 42.54 m
• B. 36.88 m
• C. 46.78 m
• D. 32.94 m

Answer 1

The correct option is B

36.88 m

Let AF = 50m be the height the of building and the swimmer moves from B to C.

In $△$ AEC

$tan{65}^{\circ }=\frac{EC}{AE}2.14=\frac{50}{AE}AE=\frac{50}{2.14}=23.36m$ (As EC = AF = 50m)

In $△$ ADB

$tan{40}^{\circ }=\frac{DB}{AD}$

$0.83=\frac{50}{AD}$

$AD=\frac{50}{0.83}=60.24m$

We can see from the figure: ED = AD - AE

ED = 60.24 - 23.36 = 36.88 m

Hence, the distance covered by the swimmer = BC = ED = 36.88 m