Question

A bird flies in a circle on a horizontal plane. An observer stands at a point on the ground. Suppose ${60}^o$ and ${30}^o$ are the maximum and the minimum angles of elevation of the bird and that they occur when the bird is at the points P and Q respectively on its path. Let $\theta$ be the angle of elevation of the bird when it is a point on the are of the circle exactly midway between P and Q . Find the numerical value of $ta{n}^2\theta$ (Assume that the observer is not inside the vertical projection of the path of the bird).

Answer 1

$\tan {60}^o=\frac{P{P}^{\prime }}{A{P}^{\prime }}=\frac{h}{d}=\sqrt{3}$ ...(1)
$\tan {30}^o=\frac{Q{Q}^{\prime }}{A{Q}^{\prime }}=\frac{h}{d+2r}=\frac{1}{\sqrt{3}}$ ...(2)
$\tan \theta =\frac{R{R}^{\prime }}{A{R}^{\prime }}=\frac{h}{\left(\sqrt{(d+r{)}^2+r^2}\right)}$
or ${\cot }^2\theta =\frac{d^2+2dr+2r^2}{h^2}$
$=\frac{d}{h}.\frac{d+2r}{h}+\frac{2r^2}{h^2}=\frac{1}{\sqrt{3}}.\sqrt{3}+2{\left(\frac{r}{h}\right)}^2$ ...(3)
Now,
$P^{\prime }{Q}^{\prime }=A{Q}^{\prime }-A{P}^{\prime }=(d+2r)-d=\sqrt{3h}-\frac{h}{\sqrt{3}}=\frac{2h}{\sqrt{3}}$ by (1) and (2)
or $2r=\frac{2h}{\sqrt{3}}$ $\therefore \frac{r^2}{h^2}=\frac{1}{3}$
Putting in (3), we get
${\cot }^2\theta =1+2.\frac{1}{3}=\frac{5}{3}$ $\therefore {\tan }^2\theta =\frac{3}{5}$