Question

An observer at O notices that the angle of elevation of the top of a tower is ${30}^o$. The line joining O to the base of the tower makes an angle of $ta{n}^{-1}\left(1/\sqrt{2}\right)$ with the North and is inclined Eastwards. The observer travels a distance of 300 metres towards the North to a point A and finds the tower to his East. The angle of elevation of the top of the tower at A is $\varphi$. Find $\varphi$ and the height of the tower.

Answer 1

$AQ=hcot\varphi ,PQ=hcot{30}^o$
Also, $tan\alpha =\frac{1}{\sqrt{2}}sin\alpha =\frac{1}{\sqrt{3}}$
or $\frac{AQ}{OQ}=sin\alpha =\frac{1}{\sqrt{3}}$ from $\Delta OAQ$
or $\frac{hcot\varphi }{hcot{30}^o}=\frac{1}{\sqrt{3}}$
$\therefore cot\varphi =1or\varphi ={45}^o$
$h=AQtan{45}^o=AQ$
or $h=OAtan\alpha =300\frac{1}{\sqrt{2}}=150\sqrt{2}$ from ground $\Delta OAQ$.