Question

The angle of elevation of the top of the building from the foot of the tower from the foot of the building is ${60}^o$. If the tower is $50m$ high, find the height of the building.

• A. $16\frac{2}{3}m$
• B. $18m$
• C. $12\frac{1}{2}m$
• D. $16m$
  

Answer 1

Answer: (A)

$16\frac{2}{3}m$

$PQ=50$meters is the height of the tower. Let $AB=h$ meters be the height of the building from the foot of the tower $={30}^o$, i.e., $\angle AQB={30}^o$.
Angle of elevation of the top of the tower from the foot of the building $={60}^o$
From $△AQB$
$\frac{h}{BQ}=\tan {30}^o=\frac{1}{\sqrt{3}}$
$\Rightarrow =h\sqrt{3}.........\left(1\right)$
From $△PBQ$
$\frac{50}{BQ}=\tan {60}^o=\sqrt{3}$
$\Rightarrow BQ=\frac{50}{\sqrt{3}}.......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have
$h\sqrt{3}=\frac{50}{\sqrt{3}}$
$\Rightarrow$ $h=\frac{50}{3}m$, i.e., $h=16\frac{2}{3}m$