ข้อสอบคณิต 2 - มีนาคม 2563

$$\begin{gathered} \mathrm{จากโจทย์} \\ \mathrm{นำข้อมูลมาวาดรูป} \mathrm{จะได้} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} A+B = 90° \\ \sin A = \cos B \\ \sin B = \cos A \\ \mathrm{จะได้} c^2\sin A = 3 \dots \overline{)1} \\ c^2\cos A = 3\sqrt{3} \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \frac{\overline{)1}}{\overline{)2}} ; \frac{c^2\sin A}{c^2\cos A} = \frac{3}{3\sqrt{3}} \\ \tan A = \frac{1}{\sqrt{3}} \\ A = 30 \mathrm{แทน} \overline{)1} \end{gathered}$$

$$\begin{gathered} c^2\sin 30° = 3 \\ c^2\left(\frac{1}{2}\right) = 3 \\ c = \sqrt{6} \end{gathered}$$

$$\begin{gathered} \mathrm{พท}. △ABC = \frac{1}{2}ab = \frac{1}{2}\left(\sqrt{6}\sin 30°\right)\left(\sqrt{6}\cos 30°\right) \\ = \frac{3\sqrt{3}}{4} \to \mathrm{ตอบ} 3 \end{gathered}$$