Question

The elevation of the toer at a station A view North of it is alpha and station view West of A is beta . Prove that the height of the tower is
$\frac{(ABsin\alpha .sin\beta )}{\sqrt{(si{n}^2\alpha -si{n}^2\beta )}}$.

Answer 1

Let O be the point where the tower stands.
Let A be the point due north of it and B the point due east of A.
From A the angle of elevation of the tower is $\alpha$
$\therefore OA/h=cot\alpha \Rightarrow OA=hcot\alpha$ ..... (1)
From B the angle of elevation of the tower is \beta .
$\therefore cot\beta =OB/h\Rightarrow OB=hcot\beta$ ....(2)
consider $\Delta OAB$
$=h^{2}co{t}^2\beta -h^{2}co{t}^2\alpha$
$h^2\left(\frac{co{s}^2\beta }{si{n}^2\beta -co{s}^2\alpha si{n}^2\alpha }\right)$
$h^2\frac{(si{n}^2\alpha co{s}^2\beta -co{s}^2\alpha si{n}^2\beta )}{si{n}^2\alpha si{n}^2\beta }$
$=h^2\left(si{n}^2\alpha \right(1-si{n}^2\beta \left)\right)-\frac{(1-si{n}^2\alpha )si{n}^2\beta }{si{n}^2\alpha si{n}^2\beta }.$
$A{B}^2=h^2\frac{(si{n}^2\alpha -si{n}^2\beta )}{si{n}^2\alpha si{n}^2\beta }$
$h^2=\frac{A{B}^{2}si{n}^2\alpha si{n}^2\beta }{(si{n}^2\alpha -si{n}^2\beta )}$
$\therefore h=AB\frac{sin\alpha sin\beta }{(si{n}^2\alpha -si{n}^2\beta )}$