Question

The angle of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are 52° and 45° respectively. Find the distance AB, to the nearest metre.

Answer 1

In ΔADC,

tan 52° = ${\displaystyle \frac{\text{DC}}{\text{AC}}=\frac{100}{\text{AC}}}$

${\displaystyle \Rightarrow }$ 1.2799 = ${\displaystyle \frac{100}{\text{AC}}}$   ...(From table)

${\displaystyle \Rightarrow }$ AC = ${\displaystyle \frac{100}{1.2799}}$

${\displaystyle \Rightarrow }$ AC = 78.13 m

In ΔBCD,

tan 45° = ${\displaystyle \frac{\text{CD}}{\text{BC}}}$

${\displaystyle \Rightarrow }$ 1 = ${\displaystyle \frac{100}{\text{BC}}}$

BC = 100 m

∴ AB = AC + BC

= 78.13 + 100

= 178.13 m

Hence, the distance AB is 178 m ...(approx)