Ex. 19. In triangle $ABC$, points $M$ and $N$ are chosen on sides $AB$ and $AC$. It is known that $\angle ABC=70^{\circ}, \angle ACB=50^{\circ}, \angle ABN=20^{\circ}$, $\angle ACM=10^{\circ}$. Find $\angle NMC$.

![](https://cdn.mathpix.com/cropped/2024_05_21_91ff4e46083e03d62f1eg-05.jpg?height=457&width=554&top_left_y=1573&top_left_x=1162)

## Generalized Sine Theorem

The length of a chord is equal to the product of the diameter of the circle and the sine of half the angular length of the arc whose ends the chord connects

$$
d=2 R \sin \frac{\varphi}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_91ff4e46083e03d62f1eg-06.jpg?height=226&width=240&top_left_y=361&top_left_x=1436)
Ex. 19. Answer: $\angle N M C=30^{\circ}$. Solution. Construct $\angle B C D$, equal to $\angle M C A$, equal to $10^{\circ}$. Lay off $D P$, equal to $A N$. Draw $P N$ and $P M . \quad \triangle C P N-$ is equilateral, therefore, $P N=C N$.

$\triangle C B N-$ is isosceles $\Rightarrow P M=B N=C N$. Hence, $P N=P M$.

Thus, $\angle P M N=80^{\circ}$. But $\angle M P=50^{\circ}$.

Therefore, $\angle N M C=80^{\circ}-50^{\circ}=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_91ff4e46083e03d62f1eg-31.jpg?height=374&width=472&top_left_y=715&top_left_x=1226)