Answer
(a). $h$ = $2\tan \theta$
(b). $ \theta$ = $\tan^{-1}(\frac{h}{2})$
Work Step by Step
At any given time, the observer, shuttle and launch pad will form a right triangle, right angled at launch pad.
Therefore, at any time-
Tangent of angle of elevation $\theta$ = height 'h' of the shuttle $\div$ distance of the observer from the launch pad
(a).
$\tan \theta$ = $\frac{h}{2}$
i.e. $h$ = $2\tan \theta$
(b).
$\tan \theta$ = $\frac{h}{2}$
$ \theta$ = $\tan^{-1}(\frac{h}{2})$
