Question

Fill in the blanks
The angle of elevation of a tower at a point $A$ due North of it is ${30}^o$ and at another point due East of $A$ is ${18}^o$. If $AB=\alpha$, the height of the tower is __________.

Answer 1

$AB=a$, $\alpha ={30}^o$ and $\beta ={18}^o$
$\therefore h=\frac{AB\sin \alpha \sin \beta }{\sqrt{({\sin }^2\alpha -{\sin }^2\beta )}}$
$\sin \alpha =\sin 30=\frac{1}{2}$,
$\sin \beta =\sin 18=\frac{\sqrt{5}-1}{4}$
$\therefore {\sin }^2\alpha -{\sin }^2\beta =\frac{1}{4}=\frac{6-2\sqrt{5}}{16}=\frac{\sqrt{5}-1}{8}$
and $\sin \alpha \sin \beta =\frac{\sqrt{5}-1}{8}$
$\therefore h=a\frac{\sqrt{5}-1}{8}.\sqrt{\left(\frac{8}{\sqrt{\left(5\right)}-1}\right)}=a\sqrt{\left(\frac{\sqrt{5}-1}{8}\right)}$
$a=\sqrt{(\frac{\sqrt{5}-1}{8}.\frac{\sqrt{5}+1}{\sqrt{\left(5\right)}+1})}$
$a=\sqrt{\left(\frac{4}{8[\sqrt{\left(5\right)}+1]}\right)}$
$a=\sqrt{\left(\frac{1}{2[\sqrt{\left(5\right)}+2]}\right)}=\frac{a}{\sqrt{[2+2\sqrt{\left(5\right)}]}}$