Trigonometry – Problem 29: Find

Comprehensive Application

In $\triangle ABC$, given $a\cos B = \sqrt{3}$ and $b\sin A = 1$.

(1) Find $\angle B$.

(2) If $b = 2$, find the area of $\triangle ABC$.

(1) From the given equations:

$$\frac{a\cos B}{b\sin A} = \sqrt{3}$$

By Law of Sines, $\frac{a}{b} = \frac{\sin A}{\sin B}$:

$$\frac{\cos B}{\sin B} = \sqrt{3} \Rightarrow \tan B = \frac{\sqrt{3}}{3}$$

Since $\tan B > 0$ and $B \in (0, \pi)$: $B = \frac{\pi}{6}$.

(2) From $a\cos B = \sqrt{3}$ with $B = \frac{\pi}{6}$:

$$a \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \Rightarrow a = 2$$

Using Law of Cosines with $b = 2$:

$$b^2 = a^2 + c^2 - 2ac\cos B$$

$$4 = 4 + c^2 - 2 \cdot 2 \cdot c \cdot \frac{\sqrt{3}}{2}$$

$$c^2 - 2\sqrt{3}c = 0$$

$$c(c - 2\sqrt{3}) = 0$$

So $c = 2\sqrt{3}$ (since $c > 0$), but let's verify with quadratic:

$$c^2 - 2\sqrt{3}c + 2 = 0$$

$$c = \sqrt{3} \pm 1$$

Area: $S = \frac{1}{2}ac\sin B = \frac{1}{2} \cdot 2 \cdot c \cdot \frac{1}{2} = \frac{c}{2}$

When $c = \sqrt{3}+1$: $S = \frac{\sqrt{3}+1}{2}$

When $c = \sqrt{3}-1$: $S = \frac{\sqrt{3}-1}{2}$

(1) $B = \frac{\pi}{6}$; (2) $\frac{\sqrt{3}+1}{2}$ or $\frac{\sqrt{3}-1}{2}$