Question

Triangle $ABC$ is isosceles with $AC=BC$ and $\angle ACB={106}^{\circ }.$ Point $M$ is in the interior of the triangle so that $\angle MAC=7^{\circ }$ and $\angle MCA={23}^{\circ }.$ Find the number of degrees in $\angle CMB.$
(correct answer + 5, wrong answer 0)

Answer 1

From the givens,
$\angle AMC={150}^{\circ },\angle MCB={83}^{\circ }.$
If we define $\angle CMB=\theta ,$
then $\angle CBM={97}^{\circ }-\theta .$
Applying sine law to $△AMC$ and $△BMC,$
$\frac{\sin {150}^{\circ }}{\sin 7^{\circ }}=\frac{AC}{CM}=\frac{BC}{CM}=\frac{\sin \theta }{\sin (97-\theta )}$
$\Rightarrow \frac{1}{2}\cos (7^{\circ }-\theta )=\sin 7^{\circ }\sin \theta$
$\Rightarrow \cos 7^{\circ }\cos \theta +\sin 7^{\circ }\sin \theta =2\sin 7^{\circ }\sin \theta$
$\Rightarrow \cos 7^{\circ }\cos \theta =\sin 7^{\circ }\sin \theta$
$\Rightarrow \tan 7^{\circ }=\cot \theta$
Since $0^{\circ }<\theta <{180}^{\circ },$
we must have $\theta ={83}^{\circ }.$