Question

A straight pillar PQ stands at a point P, the points A and B are situated due south and east of P respectively. M is the mid-point of AB. PAM is an equilateral triangle and N is the foot of the perpendicular from P on AB. Suppose AN = 20 metre and the angle of elevation of the top of the pillar at N is $ta{n}^{-1}2$. Find the height of the pillar and the angles of elevation of its top at A and B.

Answer 1

PQ is a vertical pillar. The points A and B are respectively due south and due east of the point P so that $\angle APB={90}^o$, M is the mid-point of AB and N is the foot of perpendicular from P on AB where AN = 20 metres.
N is mid-point of AM as $\Delta PAM$ is equilateral
$\therefore AM=2AN=40$ ...(1)
$AB=2AM=80$ ...(2)
Also $\angle PAM=\angle MPA={60}^o$, since the $\Delta PAM$ is equilateral. Elevation of Q as seen from N is $\alpha$ where $tan\alpha =2.$
If PQ = h, then
$PN=hcot\alpha =ANtan{60}^o=20\sqrt{3}$
$h.\left(\frac{1}{2}\right)=20\sqrt{3}$ so that $h=40\sqrt{3}$.
To find the elevations of A and B, we first find PA and PB. We have
$PA=AM=40$ by (1) as $\Delta PAM$ is equilateral
$\therefore PB=\sqrt{A{B}^2-P{A}^2}=\sqrt{{80}^2-{40}^2}$
$=40\sqrt{3}$ metre. by (1) and (2)
$tan\beta =\frac{PQ}{PA}=\frac{40\sqrt{3}}{40}=\sqrt{3}$ so that $\beta ={60}^o$
$tan\gamma =\frac{PQ}{PB}=\frac{40\sqrt{3}}{40\sqrt{\left(3\right)}}=1$ and so $\gamma ={45}^o$