Finding the angles of $\triangle ABC$

Question

In a $\triangle ABC$, from vertex $C$, the median to $AB$, the angle bisector of $\angle BCA$ and the perpendicular to $AB$ divides angle $\angle BCA$ into four equal parts.

The task is to compute angles in $\triangle ABC$.

Thanks for any help.

Answer 1

Applying sine law to triangle ACD, we have $\dfrac {s}{t} = \dfrac {\sin (90 – 3x)}{\sin x} = \dfrac {\cos 3x}{\sin x}$

Applying sine law to triangle BCD, we have $\dfrac {s}{t} = \dfrac {\sin (90 – x)}{\sin 3x} = \dfrac {\cos x}{\sin 3x}$

Then $\sin 3x \cos3x – \sin x \cos x = 0$

By Wolframalpha, we have x = $\dfrac {\pi}{2} - \dfrac {3 \pi}{8} radian = 22.5^0$ (an answer that matches @Lucian 's)

[Note: the trig equation can also be solved as $2 \sin 3x \cos3x = 2\sin x \cos x$. Then, $\sin 6x = \sin 2x$. This means $x = 0$ or $6x = \pi - 2x$; and this further yields $x = \dfrac {\pi}{8}$.]

Result follows.