102302 topics:

[Law of Cosines]

[ Law of Sines ]

A circle intersects the sides of angle $B A C$ at points $B, N, M$ and $C$, point $N$ is between $A$ and $B$, point $M$ is between $A$ and $C$. The measures of angles $A C B$ and $B M C$ are $\frac{\pi}{3}$ and $\frac{\pi}{4}$ respectively, $B N=2 M N$. What is the measure of angle $B A C$?
Apply the Law of Cosines and the Law of Sines to triangle BMN.

## Solution

Let $M N=x, B N=2 x, \angle A B M=\beta$. Since quadrilateral $B C M N$ is inscribed in a circle, then

$$
\angle B N M=\pi-\angle B C M=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}
$$

Applying the Law of Cosines and the Law of Sines to triangle $B M N$, we get:

$$
\begin{gathered}
B M=\sqrt{B N^{2}+M N^{2}-2 \cdot B N \cdot M N \cdot \cos \angle B N M}=\sqrt{4 x^{2}+x^{2}-2 \cdot 2 x \cdot x \cdot \cos \frac{2 \pi}{3}}=x \sqrt{7}, \\
\frac{M N}{\sin \beta}=\frac{B M}{\sin \frac{2 \pi}{3}}
\end{gathered}
$$

from which we find that

$$
\sin \beta=\frac{M N}{B M} \cdot \sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2 \sqrt{7}}
$$

Since $\beta<\frac{\pi}{2}$ (in triangle $M B N$ there is an obtuse angle $M N B$), then $\beta=\arcsin \frac{\sqrt{3}}{2 \sqrt{7}}$. By the theorem of the exterior angle of a triangle,

$$
\angle B A C=\angle B M C-\angle A B M=\frac{\pi}{4}-\beta=\frac{\pi}{4}-\arcsin \frac{\sqrt{3}}{2 \sqrt{7}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-26.jpg?height=563&width=972&top_left_y=2308&top_left_x=543)
$\angle B A C=\frac{\pi}{4}-\arcsin \frac{\sqrt{3}}{2 \sqrt{7}}=\operatorname{arctg} \frac{\sqrt{3}}{2}-\frac{\pi}{12}$.

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