Q16 (c) (ii)

(16)(c)(ii)
$\tan(\alpha_1+\alpha_2)=\frac{\tan\alpha_1+\tan\alpha_2}{1-\tan\alpha_1\tan\alpha_2}\quad (1)$

Note: $\tan\alpha_1+\tan\alpha_2=\dfrac{2V^2}{ga}\quad (2)$

$\tan\alpha_1\tan\alpha_2=\frac{V^4-(V^4-g^2a^2-2gV^2b)}{g^2a^2}$
$=\frac{g^2a^2+2gV^2b}{g^2a^2}$
$=1+\frac{2V^2b}{ga^2}\quad (3)$

Sub. (2) & (3) in (1):
$\tan(\alpha_1+\alpha_2)=\frac{\frac{2V^2}{ga}}{1-\left(1+\frac{2V^2b}{ga^2}\right)}$
$=\frac{2V^2}{ga}\div\frac{-2V^2b}{ga^2}$
$=\frac{2V^2}{ga}\times\frac{-ga^2}{2V^2b}$
$=-\frac{a}{b}$

As $\tan\theta=\frac{b}{a}$,
$\tan(90^\circ+\theta)=\tan[180^\circ-(90^\circ-\theta)]$
$=-\tan(90^\circ-\theta)$
$=-\frac{1}{\tan\theta}$
$=-\frac{a}{b}$

$\therefore\tan(\alpha_1+\alpha_2)=\tan(90^\circ+\theta)$

$\therefore\alpha_1+\alpha_2=90^\circ+\theta$

$\therefore\alpha_1=90^\circ-\theta-\alpha_2$

$\angle BOA=\angle COY$