Question

The angle elevation of the top $Q$ of a vertical tower $PQ$ from a point $X$ on the ground is $60$ degree.From a point $Y$,$40$m vertically above $X$,the angles of elevation of the top $Q$ of the tower is ${45}^{\circ }$.Find the height of the tower $PQ$ and the distance $PX.$(Use $\sqrt{3}=1.732$)

Answer 1

Given:The angle of elevation of the top $Q$ of a vertical tower from the $PQ$ from a point $X$ on the ground is ${60}^{\circ }$.From a point $Y$,$40$m vertically above $X$,the angle of elevation of the top $Q$ of the tower is ${45}^{\circ }$

$MP=XY=40$m

$\therefore QM=h-40$

In rightangled $△QMY$,

$\tan {45}^{\circ }=\frac{QM}{MY}=1=\frac{h-40}{PX}$ since $MY=PX$

$\therefore PX=h-40$ .....$\left(1\right)$

$\tan {60}^{\circ }=\frac{QP}{PX}$

$\Rightarrow \sqrt{3}=\frac{QP}{PX}$

$\therefore PX=\frac{h}{\sqrt{3}}$ ........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$h-40=\frac{h}{\sqrt{3}}$

$\therefore h\sqrt{3}-40\sqrt{3}=h$

$\Rightarrow h(\sqrt{3}-1)=40\sqrt{3}$

$\Rightarrow h=\frac{40\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\Rightarrow h=\frac{40\sqrt{3}(\sqrt{3}+1)}{3-1}$

$\Rightarrow h=\frac{40\sqrt{3}(\sqrt{3}+1)}{2}$

$\therefore h=20(3+\sqrt{3})=20(3+1.73)=20\times 4.73=94.6‬$m