高卒認定試験 数学 H28-2 大問5(5)解説+解答

解説:

正弦定理:

$$ \frac{a}{\sin{\angle{A}}} = \frac{b}{\sin{\angle{B}}} =2R $$


三角比の値:

$$ \begin{eqnarray}
\sin{\angle{30^\circ}}&=&\frac{1}{2}\\
\sin{\angle{45^\circ}}&=&\frac{1}{\sqrt{2}}\\
\end{eqnarray} $$



$$ \begin{eqnarray}
\frac{a}{\sin{A}} &=& \frac{b}{\sin{B}} \\
\frac{CB}{\sin{\angle{A}}} &=& \frac{AB}{\sin{\angle{C}}} \\
\frac{3cm}{\sin{30^\circ}} &=& \frac{AB}{\sin{45^\circ}} \\
\frac{3cm}{\frac{1}{2}} &=& \frac{AB}{\frac{1}{\sqrt{2}}} \\
&&左辺の分母分子に2を掛け、右辺の分母分子に\sqrt{2}を掛けると、\\
\frac{3cm \times{2}}{\frac{1}{2} \times {2}} &=& \frac{AB \times {\sqrt{2}}}{\frac{1 \times{\sqrt{2}}}{\sqrt{2}}} \\
3cm \times 2 &=& AB \times \sqrt{2} \\
&& 左辺と右辺を入れ替えて\\
AB \times \sqrt{2}&=& 3cm \times 2 \\
AB &=& \frac{6cm} {\sqrt{2}} \\
&&分母の有理化(分母分子に\sqrt{2}を掛ける)\\
AB &=& \frac{6cm \times \sqrt{2}}{\sqrt{2} \times  \sqrt{2}} \\
AB &=&  \frac{6cm \times \sqrt{2}}{2}\\
AB &=& 3\sqrt{2}cm \\
\end{eqnarray} $$