ข้อสอบ PAT 1 - ตุลาคม 2559

$$\begin{gathered} \left(\mathrm{ก}\right) \mathrm{สูตรกฎของ} \mathrm{cosine} \hspace{0.17em}\hspace{0.17em} {\mathrm{a}}^2={\mathrm{b}}^2+{\mathrm{c}}^2-2\mathrm{bc} \mathrm{cosA} \\ {a}^2={\left(\frac{1}{\sqrt{6}+\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{6}-\sqrt{2}}\right)}^2-2\left(\frac{1}{\sqrt{6}+\sqrt{2}}\right)\left(\frac{1}{\sqrt{6}-\sqrt{2}}\right)\cos 60° \\ =\frac{1}{6+2\sqrt{12}+2}+\frac{1}{6-2\sqrt{12}+2}-2\left(\frac{1}{6-2}\right)\frac{1}{2} \\ =\frac{1}{8+4\sqrt{3}}+\frac{1}{8-4\sqrt{3}}-\frac{1}{4} \\ =\frac{\left(8-\cancel{4\sqrt{3}}\right)+8+\cancel{4\sqrt{3}}}{{\left(8\right)}^2-{\left(4\sqrt{3}\right)}^2}-\frac{1}{4} \end{gathered}$$

$$\begin{gathered} =\frac{16}{64-48}-\frac{1}{4} \\ =1-\frac{1}{4} \\ =\frac{3}{4} \\ \mathrm{a}=\frac{\sqrt{3}}{2} \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ข}\right) \mathrm{สูตรกฎของ} \sin ; \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\ \frac{{\displaystyle \frac{\sqrt{3}}{2}}}{\sin 60°}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \hspace{0.17em}\hspace{0.17em}; \sin 60°=\frac{\sqrt{3}}{2} \\ 1=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\ \mathrm{จะได้} 1=\frac{\mathrm{b}}{\sin \mathrm{B}} \mathrm{และ} 1=\frac{\mathrm{c}}{\sin \mathrm{C}} \\ \sin \mathrm{B}=\mathrm{b} \sin \mathrm{C}=\mathrm{c} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} {\sin }^2\mathrm{B}+{\sin }^2\mathrm{C}={\mathrm{b}}^2+{\mathrm{c}}^2 ; {\mathrm{b}}^2+{\mathrm{c}}^2=1 \mathrm{ได้จากข้อ} \left(\mathrm{ก}\right) \\ =1 \mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ถูก} \\ \left(\mathrm{ค}\right) \mathrm{จาก} \left(\mathrm{ข}\right) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sin \mathrm{B}=\mathrm{b}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{และ}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sin \mathrm{C}=\mathrm{c} \\ \mathrm{แสดงว่า} \sin \mathrm{B}+\sin \mathrm{C}=\mathrm{b}+\mathrm{c} \\ =\frac{1}{\sqrt{6}+\sqrt{2}}+\frac{1}{\sqrt{6}-\sqrt{2}} \end{gathered}$$
$$\begin{gathered} =\frac{1}{\sqrt{6}+\sqrt{2}}+\frac{1}{\sqrt{6}-\sqrt{2}} \\ =\frac{\left(\sqrt{6}-\cancel{\sqrt{2}}\right)+\left(\sqrt{6}+\cancel{\sqrt{2}}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)} \\ =\frac{2\sqrt{6}}{6-2} \\ =\frac{\sqrt{6}}{2} \mathrm{ข้อ} \left(\mathrm{ค}\right) \mathrm{ผิด} \end{gathered}$$