Example 34. In an isosceles triangle $A B C$, where the angle $C$ at the vertex is $100^{\circ}$, two rays are constructed: one starting from point $A$ at an angle of $30^{\circ}$ to ray $A B$, and the other starting from point $B$ at an angle of $20^{\circ}$ to ray $B A$. The constructed rays intersect at point $M$, which lies within triangle $A B C$. Find the angles $A C M$ and $B C M$.
Solution (Fig. 38). Let's provide an algebraic solution.

Construct $[C M],\left[M A_{1}\right] \perp(B C),\left[M B_{1}\right] \perp(A C),\left[M C_{1}\right] \perp(A B)$. Let $\widehat{A C M}=x$, then $\widehat{B C M}=100^{\circ}-x$. Introduce an auxiliary parameter, setting $|C M|=k$, and calculate $\left|M C_{1}\right|$ in two ways, i.e., use $\left|M C_{1}\right|$ as the reference element.

From triangles $C M B_{1}$ and $C M A_{1}$, we have respectively:

$$
\left|M B_{1}\right|=k \sin x,\left|M A_{1}\right|=k \sin \left(100^{\circ}-x\right) .
$$

Since angle $C$ is obtuse and triangle $A B C$ is isosceles, then

$$
\widehat{A}=\widehat{B}=\frac{180^{\circ}-100^{\circ}}{2}=40^{\circ}
$$

Since $\widehat{M A} C_{1}=30^{\circ}$ and $\widehat{M B C_{1}}=20^{\circ}$, then $\widehat{C A M}=10^{\circ}$ and $\widehat{C B M}=$ $=20^{\circ}$.

Thus, in right triangles $A_{1} B M$ and $C_{1} B M[B M]-$ the common hypotenuse and $\widehat{C B M}=\widehat{A B M}$, i.e., $\triangle A_{1} B M \cong \triangle C_{1} B M$, and then $\left[M C_{1}\right] \cong\left[A_{1} M\right]$, or

$$
\left|M C_{1}\right|=k \sin \left(100^{\circ}-x\right)
$$

From $\triangle A B_{1} M$ we have:

$$
|A M|=\frac{\left|M B_{1}\right|}{\sin 10^{\circ}}=\frac{k \sin x}{\sin 10^{\circ}}
$$

and, therefore,

$$
\left|M C_{1}\right|=|A M| \cdot \sin 30^{\circ}
$$

hence

$$
\left|M C_{1}\right|=\frac{k \sin x}{2 \sin 10^{\circ}}
$$

Comparing the values of $\left|M C_{1}\right|$ from (34.1) and (34.2), we arrive at the equation

$$
k \sin \left(100^{\circ}-x\right)=\frac{k \sin x}{2 \sin 10^{\circ}}
$$

or

$$
2 \sin 10^{\circ} \sin \left(100^{\circ}-x\right)=\sin x
$$

from which we find: $x=20^{\circ}$.

Conclusion. Since point $M$ belongs to $\triangle A B C$, then

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-057.jpg?height=171&width=868&top_left_y=1456&top_left_x=451)

i.e., $0^{\circ}<x<100^{\circ}$. The found value of $x$ satisfies this system. Therefore, $\widehat{A C M}=$

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-057.jpg?height=614&width=682&top_left_y=1755&top_left_x=350)

Fig. 39 $=20^{\circ}, \widehat{B C M}=80^{\circ}$.