7.7. Consider the triangle $A B C$ with $A B=A C$ and $m(\angle B)>30^{\circ}$. Inside the triangle, we have a point $M$ such that $m(\angle M B C)=30^{\circ}$ and $m(\angle M A B)=\frac{3}{4} m(\angle B A C)$. Determine $m(\angle A M C)$.
## Solution:

We construct the bisector $(A D), D \in(B C)$. Let $\{P\}=A D \cap B M$. Since $(A D)$ is also an altitude, we deduce that triangle $B P C$ is isosceles, so $m(\angle B P D)=m(\angle D P C)=60^{\circ}$. Therefore,

$m(\angle A P M)=m(\angle C P M)=60^{\circ}$.

From $m(\angle M A B)=\frac{3}{4} m(\angle B A C)$ we get $m(\angle P A M)=m(\angle M A C)=\frac{1}{4} m(\angle B A C)$. Thus, $A M$ and $P M$ are bisectors in triangle $A P C$. It follows that $C M$ is also a bisector, i.e., $m(\angle A C M)=$ $m(\angle P C M)$.

In triangle $M A C$ we obtain $m(\angle A M C)=180^{\circ}-[m(\angle M A C)+m(\angle M C A)]=180^{\circ}-$ $\frac{1}{2}[m(\angle P A C)+m(\angle P C A)]=180^{\circ}-\frac{1}{2}\left[180^{\circ}-120^{\circ}\right]=180^{\circ}-30^{\circ}=150^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_bc35319069f73311db41g-2.jpg?height=634&width=620&top_left_y=765&top_left_x=798)