ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$\mathrm{สูตรตรีโกณ} \sin \left(A-B\right) = \sin A\cos B-\cos A\sin B$

$$\begin{gathered} \mathrm{จากโจทย์} ABC \mathrm{เป็นรูปสามเหลี่ยม} \\ \mathrm{จะได้} A+B+C = 180° \\ A+B+30° = 180° \\ A = 150°-B \to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{กฎของ} \sin \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \\ \frac{\sin A}{a} = \frac{\sin B}{b} \\ \frac{\sin \left(150-B\right)}{a} = \frac{\sin B}{a\left(\sqrt{3}-1\right)} \\ sin\left(150-B\right) = \frac{\sin B}{\sqrt{3}-1} \end{gathered}$$

$$\begin{gathered} \sin 150\cos B-\cos 150\sin B = \frac{\sin B}{\sqrt{3}-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}+1}\right) \\ \sin 30\cos B-\left(-\cos 30\right)\sin B = \frac{\sin B\left(\sqrt{3}+1\right)}{3-1} \\ \frac{1}{2}\cos B+\frac{\sqrt{3}}{2}\sin B = \frac{\left(\sqrt{3}+1\right)}{2}\sin B \end{gathered}$$

$$\begin{gathered} \cos B+\sqrt{3}\sin B = \left(\sqrt{3}+1\right)\sin B \\ \cos B = \sin B \\ \tan B = 1 \\ B = 45° \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \sin 3B = \sin 3\left(45°\right) \\ = \sin 135° \\ = \sin 45° = \frac{\sqrt{2}}{2} \end{gathered}$$