ข้อสอบคณิต O-NET - 2553

$\mathrm{จากโจทย์} \mathrm{นำมาวาดรูป}$

$$\begin{gathered} \mathrm{กำหนดให้} \mathrm{ด้าน} BC = x \\ \mathrm{จะได้} \cos 60° = \frac{BC}{AB} \\ \frac{1}{2} = \frac{BC}{AB} \\ AB = 2BC \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \tan 60° = \frac{AC}{BC} \\ \sqrt{3} = \frac{AC}{x} \\ AC = \sqrt{3}x \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \left(2x\right)+\left(x\right)+\left(\sqrt{3}x\right) = 3-\sqrt{3} \\ \left(3+\sqrt{3}\right)x = 3-\sqrt{3} \\ x = \frac{3-\sqrt{3}}{3+\sqrt{3}}\times \frac{\left(3-\sqrt{3}\right)}{\left(3-\sqrt{3}\right)} \end{gathered}$$

$$\begin{gathered} = \frac{{\left(3-\sqrt{3}\right)}^2}{9-3} \\ = \frac{9-2\left(3\right)\left(\sqrt{3}\right)+3}{6} \\ = \frac{12-6\sqrt{3}}{6} \\ = 2-\sqrt{3} \end{gathered}$$