Question

If $\beta$ and $\alpha$ are +ive angles less than two right angles and $cos\alpha +cos(\alpha +\beta )+cos(\alpha +\beta +\gamma )=0$
and $sin\alpha +sin(\alpha +\beta )+sin(\alpha +\beta +\gamma )=0$
then $\beta =\gamma =2\pi /3$

Answer 1

Transpose the last term and then square and add
1 + 1 +2 cos ($\alpha +\beta -\alpha$) = 1
$\therefore cos\beta =-\frac{1}{2}\therefore \beta =\frac{2\pi }{3}$
Again 2 cos $(\alpha +\frac{\beta }{2})cos\frac{\beta }{2}=-cos(\alpha +\beta +\gamma )$
2 sin $(\alpha +\frac{\beta }{2})cos\frac{\beta }{2}=-sin(\alpha +\beta +\gamma )$
$\therefore tan(\alpha +\frac{\beta }{2})=tan(\alpha +\beta +\gamma )$
$\therefore \alpha +\beta +\gamma =\alpha +\frac{\beta }{2}=\pi (\alpha +\frac{\beta }{2})$
The first gives as $\gamma =-\frac{\beta }{2}=-\frac{\pi }{3}$ rejected as $\gamma$ is +ive
From 2nd, $\gamma =\pi -\frac{\beta }{2}=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$