Trigonometry – Problem 22: find angle

Law of Sines Side-Angle Conversion

In $\triangle ABC$, let sides opposite to angles $A, B, C$ be $a, b, c$. Given $\sin(B+C) + \sin A = \frac{3}{2}$ and $b = \sqrt{3}c$, find angle $C$.

In a triangle, $\sin A = \sin(B+C)$, so:

$$\sin(B+C) + \sin A = 2\sin A = \frac{3}{2}$$

$$\sin A = \frac{3}{4}$$

Since $b = \sqrt{3}c$, by Law of Sines: $\sin B = \sqrt{3}\sin C$.

Using $\sin A = \sin(B+C)$:

$$\sin B\cos C + \cos B\sin C = \frac{3}{4}$$

$$\sqrt{3}\sin C\cos C + \cos B\sin C = \frac{3}{4}$$

This gives $\sin 2C = \frac{\sqrt{3}}{2}$, so $2C = \frac{\pi}{3}$ or $\frac{2\pi}{3}$.

If $C = \frac{\pi}{3}$, then $\sin B = \sqrt{3}\sin\frac{\pi}{3} = \frac{3}{2} > 1$ (impossible).

Therefore $C = \frac{\pi}{6}$.

$\frac{\pi}{6}$