Question

A vertical tower $PQ$ subtends the same angle ${30}^o$ at each of two places $A$ and $B,60m$ apart on the ground, $AB$ subtends an angle ${120}^o$ at the foot of the tower. If $h$ is the height of the tower, then $9h^2+h+1$ is equal to

• A. $3121$
• B. $3621$
• C. $3600$
• D. $3681$

Answer 1

The correct option is B $3621$

$\angle PAQ=\angle PBQ={30}^o$

$\tan \angle PQ=\frac{h}{AQ}$

$\tan {30}^o=\frac{h}{AQ}$ so $AQ=h\sqrt{3}$ & $BQ=h\sqrt{3}$

So ABQ is an isosceles triangle

So $\angle QAB=\angle QBA=\frac{1}{2}({180}^o-{120}^o)={30}^o$

In $\Delta ADQ:-\cos {30}^o=\frac{30}{AQ}$

So $AQ=\frac{30\cdot 2}{\sqrt{3}}=\frac{60}{\sqrt{3}}=20\sqrt{3}=h\sqrt{3}$

So $h=20$

So $9h^2+h+1$

$\Rightarrow 9\left(400\right)+20+1$

$\Rightarrow 3600+20+1=3621$.