Question

Two men are on diametrically opposite side of a tower. They measure the angle of elevation of the top of the tower as${20}^0$ and ${24}^0$ respectively. If the height of the tower is $40m$, find the distance between them.
[Hint.use tables to find tan ${20}^0$ and tan${24}^0$]

Answer 1

As we know that,

$\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}$

Therefore,

In $△ABD$,

$\tan 20°=\frac{AD}{BD}$

$\Rightarrow 0.364=\frac{40}{BD}$ $(\because \tan 20°=0.364)$

$\Rightarrow BD=\frac{40}{0.364}$

$\Rightarrow BD=109.89$

In $△ACD$

$\tan 24°=\frac{AD}{CD}$

$\Rightarrow 0.445=\frac{40}{CD}$ $(\because \tan 24°=0.445)$

$\Rightarrow CD=\frac{40}{0.445}$

$\Rightarrow CD=89.89$

Therefore,

Required distance between the two men $=BD+CD=(109.89+89.89)m=199.78m$

Hence the distance between the two men is $199.78m$.