Question

A vertical tower $PQ$ subtends the same angle of ${30}^0$ at each of two places $A$ and $B$. $60m$ apart on the ground. If $AB$ students an angle of ${120}^0$ at $P$ the foot of the tower, the height of the tower is.....

• A. $20m$
• B. $15m$
• C. $25m$
• D. $22m$

Answer 1

The correct option is A $20m$

Here,

$\angle PAQ=\angle PBQ$

$\tan PAQ=\frac{h}{AQ}$

$\tan {30}^o=\frac{h}{AQ}$

$AQ=h\sqrt{3}$

And, $BQ=h\sqrt{3}$

So, $ABQ$ is an isosceles triangle.

So, $\angle QBA=\angle QAB=\frac{1}{2}({120}^o-{120}^o)={30}^o$

In $△ADQ$,

$\cos {30}^o=\frac{30}{AQ}$

So, AQ = $\frac{30}{\sqrt{3}}\times 2=20\sqrt{3}=h\sqrt{3}$

Hence, $h=20$

Hence, option (a) is correct.