## Task B-2.4.

For the angles of a triangle $\alpha, \beta$, and $\gamma$, if $\sin \alpha = 2 \sin \beta \cos \gamma$ and $\beta + \gamma = 86^{\circ}$, determine $\alpha, \beta$, and $\gamma$.
## Solution.

Let $a$ be the length of the side opposite to angle $\alpha$, and $b$ be the length of the side opposite to angle $\beta$ in the given triangle. If we write the expression $\sin \alpha = 2 \sin \beta \cos \gamma$ in the form

$$
\frac{\sin \alpha}{\sin \beta} = 2 \cos \gamma
$$

then by applying the Law of Sines and the Law of Cosines, we get

$$
\frac{a}{b} = 2 \cdot \frac{a^2 + b^2 - c^2}{2ab}
$$

or

$$
\frac{a}{b} = \frac{a^2 + b^2 - c^2}{ab}
$$

Since we are dealing with a triangle and the length of side $b \neq 0$, after multiplying by $ab$ we get

$$
a^2 = a^2 + b^2 - c^2
$$

1 point

so $b^2 - c^2 = 0$, from which it follows that $b = c$.

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Therefore, the given triangle is isosceles, and the angles opposite the sides of length $b$ and $c$ are equal.

Then $\beta = \gamma = \frac{86^\circ}{2} = 43^\circ, \alpha = 180^\circ - 86^\circ = 94^\circ$.