ข้อสอบคณิต 2 - มีนาคม 2563

$\mathrm{จากโจทย์} \mathrm{นำข้อมูลมาวาดรูป}$

$$\begin{gathered} \mathrm{จะได้} A+B = 90° \\ B = 90°-A \\ ก. \sin A = \cos \theta \\ = \cos \left(90°-A\right) \\ = \sin A \to \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} ข. \hat{A} = 2\hat{B} \\ B = 90°-2B \\ 3B = 90° \\ B = 30° \to A = 90°-30° = 60° \\ \cos A = \cos 60° = \frac{1}{2} \to \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} ค. \sin A < \tan A \\ \sin A < \frac{\sin A}{\cos A} \\ \cos A < 1 \end{gathered}$$

$$\begin{gathered} A = 30° \to \cos 30° = \frac{\sqrt{3}}{2} < 1 \\ A = 45° \to \cos 45° = \frac{\sqrt{2}}{2} < 1 \\ A = 60° \to \cos 60° = \frac{1}{2} < 1 \\ ค. \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} ง. \sin A < \sin B \\ \frac{a}{c} < \frac{b}{c} \\ a < b \to \mathrm{ถูก} \\ \mathrm{ดังนั้น} \mathrm{ตอบข้อ} 5 \end{gathered}$$