Question

The angle od elevation of the top of tower from two points $A$ and $B$ lying on the horizontal through the foot of the tower are respectively ${15}^o$ and ${30}^o$. If $A$ and $B$ are on the same side of the tower and $AB=48m$, then the height of the tower is:

Answer 1

$CD=tower=h$

$\tan 15°=\frac{h}{48+x}⟶\left(i\right)$

$\tan 30°=\frac{1}{\sqrt{3}}=\frac{h}{x}⟹h=\frac{x}{\sqrt{3}}⟹x=h\sqrt{3}$

$\tan 15°=\tan (45°-30°)=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{h}{48+x}=\frac{h}{48+h\sqrt{3}}$

$48\sqrt{3}-48+3h-h\sqrt{3}=h\sqrt{3}+h$

$h=\frac{48{(\sqrt{3}-1)}^2}{2}=24(4-2\sqrt{3})m$