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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.

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  • $\begingroup$ It's a right triangle with an angle of $\dfrac\pi8.~$ I know this by heart, because I've asked myself a somewhat similar question several months ago, namely, in what triangle do the angle bisector and the height or altitude trisect one of the two segments formed by the median on the opposite side. $($Initially, I wanted to find a triangle such that the three lines quadrisect the basis, but that is impossible, since the angle bisector is always trapped between the median and the height or altitude$).$ $\endgroup$ – Lucian Dec 3 '15 at 15:35
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    $\begingroup$ Let $\angle A = x$. Then $\angle B$ = ... = 3x - 180. The median CD divides ABC into two smaller triangles. In them, BD of $\triangle BCD$ = DA of $\triangle ACD$. Also, CD is the common side to both triangles. Applying sine law to both triangles, we might be able to get a trig equation in x. A value of x may be obtained if that equation is not going round and round. Just a suggestion and I have no time to try. $\endgroup$ – Mick Dec 3 '15 at 15:42
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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.

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