Question

From the top of a $7m$ building, the angle of elevation of the top of a cable tower is ${60}^o$ and the angle of depression of its foot is ${45}^o$. Determine the height of the tower.

• A. $5(\sqrt{2}-1)m$
• B. $9(\sqrt{3}-1)m$
• C. $4(\sqrt{5}+1)m$
• D. $7(\sqrt{3}+1)m$

Answer 1

Answer: (D)

$7(\sqrt{3}+1)m$

Let $PQ=h$ meters be the height of the cable tower. $AB=7$ meters is the height of the building $\angle PAR={60}^o$ is the angleof elevation of the top of the cable tower from the top of the building.
$\angle RAQ={45}^o$ is the angle of depression of the foot of the cable tower from the top of the building. Then $\angle AQB={45}^o$.
Now $BQ=AR=x$ meters (say)
From $△AQB$, $\frac{AB}{BQ}=\tan {45}^o$ $\Rightarrow$ $\frac{7}{x}=1$ $\Rightarrow x=7m$
Now, from $△PAR$
$\frac{PR}{AR}=\tan {60}^o$ $\Rightarrow \frac{PQ-QR}{x}=\sqrt{3}$
$\Rightarrow$ $\frac{h-7}{x}=\sqrt{3}$ $\Rightarrow \frac{h-7}{7}=\sqrt{3}$
$\Rightarrow$ $h=7(\sqrt{3}+1)$
Hence, the height of the cable tower is $7(\sqrt{3}+1)$