Question

From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of a $20m$ high building are ${45}^o$ and ${60}^o$ respectively. Find the height of the tower.

• A. $h=15m$
• B. $h=5(\sqrt{5}-1)m$
• C. $h=20(\sqrt{3}-1)m$
• D. $h=22m$

Answer 1

The correct option is C $h=20(\sqrt{3}-1)m$

$PQ=20m$ is the height of the building.
Let $PR=h$ meters be the height of the transmission tower. $P$ is the bottom and $R$ is the top of the transmission tower
$\angle POQ={45}^o$ and $\angle POQ={60}^o$
From $△OPQ$,
$\frac{PQ}{OQ}=\tan {45}^o$
$\Rightarrow \frac{20}{OQ}=1$
$\Rightarrow OQ=20m$
From $△ORQ$,
$\frac{RQ}{OQ}=\tan {60}^o$
$\Rightarrow \frac{20+h}{20}=\sqrt{3}$
($\because RQ=PQ+PR=20+h$meters and $OQ=20$meters)
$\Rightarrow 1+\frac{h}{20}=\sqrt{3}$ $\Rightarrow \frac{h}{20}=(\sqrt{3}-1)$
$\Rightarrow h=20(\sqrt{3}-1)m$