Question

A right circular cylindrical tower of height h and radius r stands on a horizontal plane. Let A be a point in the horizontal plane and PQR be the semi-circular edge of the top of the tower such that Q is the point in it nearest to A, the angles of elevation of the points P and Q from A are ${45}^o$ and ${60}^o$ respectively. Show that $\frac{h}{r}=\frac{\sqrt{3}(1+\sqrt{5})}{2}$

Answer 1

Figure is self-explanatory, we have,
Let $P^{\prime },Q^{\prime },R^{\prime }$ and $O^{\prime }$ be the projection of $P,Q,R$ and $O$ in the base of the tower.
$A{P}^{\prime }=A{R}^{\prime }=h\cot {45}^{\circ }=h$
and $A{Q}^{\prime }=h\cot {60}^{\circ }=\frac{h}{\sqrt{3}}$
Now in $\angle A{O}^{\prime }{P}^{\prime },$ we have

$A{O}^{\prime }=A{Q}^{\prime }+Q^{\prime }{O}^{\prime }=\frac{h}{\sqrt{\left(3\right)}}+r$

$O^{\prime }{P}^{\prime }=r,A{P}^{\prime }=h$ and $\angle A{O}^{\prime }{P}^{\prime }={90}^{\circ }$

Hence $O^{\prime }{A}^2+O^{\prime }{P}^2=A{P}^2$

$\therefore {(\frac{h}{\sqrt{\left(3\right)}}+r)}^2=r^2=h^2$

or $h^2-\sqrt{3}hr-3r^2=0$.

$\therefore h=\frac{\sqrt{3}r\pm \sqrt{3r^2+12r^2}}{2}$

or $\frac{h}{r}=\frac{\sqrt{3}+\sqrt{15}}{2}=\frac{\sqrt{3}(\sqrt{5}+1)}{2}$