ข้อสอบ PAT 1 - เมษายน 2557

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{มุม} \mathrm{A} \mathrm{มีขนาดเป็นสองเท่าของมุม} \mathrm{B} \\ \mathrm{จะได้} \mathrm{A}=2\mathrm{B} \\ \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{ABC} \mathrm{เป็นรูปสามเหลี่ยม} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \mathrm{A}+\mathrm{B}+\mathrm{C}=180° \\ \mathrm{C}=180°-\left(\mathrm{A}+\mathrm{B}\right) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ; \mathrm{take} \sin \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ \sin \mathrm{C}=\sin \left(\left[180-\left(\mathrm{A}+\mathrm{B}\right)\right]\right) ; \sin \left(180-\mathrm{\theta }\right)=\mathrm{sin\theta } \\ \sin \mathrm{C}=\sin \left(\mathrm{A}+\mathrm{B}\right) ; \mathrm{โดย} \mathrm{A}=2\mathrm{B} \\ \sin \mathrm{C}=\sin \left(2\mathrm{B}+\mathrm{B}\right) \\ \sin \mathrm{C}=\sin \left(3\mathrm{B}\right) \\ \mathrm{กฎของ} \sin \frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin \mathrm{C}}{\mathrm{c}} \end{gathered}$$
$$\begin{gathered} \mathrm{จะได้} \frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}} \mathrm{และ} \frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin \mathrm{C}}{\mathrm{c}} \\ \frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}} \frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin 3\mathrm{B}}{\mathrm{c}} \\ \frac{\sin 2\mathrm{B}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}} \frac{ \sin \mathrm{B}}{\mathrm{b}}=\frac{3\sin \mathrm{B}-4{\sin }^3 \mathrm{B}}{\mathrm{c}} \\ \frac{2\sin \mathrm{B} \cos \mathrm{B}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}} \frac{1}{\mathrm{b}}=\frac{3-4{\sin }^2 \mathrm{B}}{\mathrm{c}} \\ \cos \mathrm{B}=\frac{\mathrm{a}}{2\mathrm{b}}\to \boxed{1} \frac{1}{\mathrm{b}}=\frac{3-4\left(1-{\cos }^2 \mathrm{B}\right)}{\mathrm{c}} \\ \frac{1}{\mathrm{b}}=\frac{4{\cos }^2 \mathrm{B}-1}{\mathrm{c}}\to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{แทนค่า} \cos \mathrm{B} \mathrm{ลงใน} \boxed{2} \\ \mathrm{จะได้} \frac{1}{\mathrm{b}}=\frac{4{\left({\displaystyle \frac{\mathrm{a}}{2\mathrm{b}}}\right)}^2-1}{\mathrm{c}} \\ \frac{\mathrm{c}}{\mathrm{b}}=4\left(\frac{{\mathrm{a}}^2}{4{\mathrm{b}}^2}\right)-1 \\ \mathrm{cb}={\mathrm{a}}^2-{\mathrm{b}}^2 \\ \mathrm{ดังนั้น} {\mathrm{a}}^2={\mathrm{b}}^2+\mathrm{bc} \end{gathered}$$