R2年度第2回高認数学問5解説
問5
(1)【文章題(三角比)】
$\mathrm{AB} = 10 \times \cos 50^\circ$$= 10 \times 0.6428 = 6.428$
(2)【三角比】
$\sin 130^\circ = \sin (180^\circ - 130^\circ)$$= \sin 50^\circ = 0.7660$
(3)【三角比、三平方の定理】
三平方の定理より$a^2 + b^2 = c^2$$\sin \theta = \dfrac{a}{c}, \; \cos \theta = \dfrac{b}{c}$とすると
$\sin^2 \theta + \cos^2 \theta = \left( \dfrac{a}{c} \right)^2 + \left( \dfrac{b}{c} \right)^2$
$= \dfrac{a^2}{c^2} + \dfrac{b^2}{c^2} = \dfrac{a^2 + b^2}{c^2}$
$= \dfrac{c^2}{c^2} = 1$
(4)【余弦定理】
$\mathrm{BC}^2 = \mathrm{AB}^2 + \mathrm{AC}^2 - 2 \cdot \mathrm{AB} \cdot \mathrm{AC} \cdot \cos A$$= 1^2 + 3^2 - 2\cdot 1 \cdot 3 \cdot \left( -\dfrac{1}{6} \right)$
$= 1 + 9 + 1 = 11$
よって、$\mathrm{BC} = \sqrt{11}$
(5)【正弦定理】
$\dfrac{\mathrm{BC}}{\sin A} = \dfrac{\mathrm{AB}}{\sin C}$$\mathrm{BC} = \dfrac{\mathrm{AB}}{\sin C} \times \sin A$
$= \dfrac{6}{\dfrac{\sqrt{3}}{2}} \times \dfrac{\sqrt{2}}{2} = 6 \div \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{2}}{2}$
$= 6 \times \dfrac{2}{\sqrt{3}} \times \dfrac{\sqrt{2}}{2} = \dfrac{6 \times 2 \times \sqrt{2}}{\sqrt{3} \times 2}$
$= \dfrac{6\sqrt{2}}{\sqrt{3}} = \dfrac{6\sqrt{6}}{3}= 2\sqrt{6}$
ちなみに、点$\mathrm{B}$から線分$\mathrm{AC}$に垂線を引き、その垂線と線分$\mathrm{AC}$との交点を$\mathrm{H}$とすると
$\mathrm{BH} = \mathrm{AB} \times \sin A$
$= 6 \times \dfrac{\sqrt{2}}{2} = 3\sqrt{2}$
$\mathrm{BH} = \mathrm{BC} \times \sin C$
$= \mathrm{BC} \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}\mathrm{BC}$
$\dfrac{\sqrt{3}}{2}\mathrm{BC} = 3\sqrt{2}$
$\mathrm{BC} = 3\sqrt{2} \div \dfrac{\sqrt{3}}{2}$
$= 3\sqrt{2} \times \dfrac{2}{\sqrt{3}} = \dfrac{6\sqrt{2}}{\sqrt{3}}$
$= \dfrac{6\sqrt{6}}{3} = 2\sqrt{6}$
$\mathrm{BH} = \mathrm{BC} \times \sin C$
$= \mathrm{BC} \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}\mathrm{BC}$
$\dfrac{\sqrt{3}}{2}\mathrm{BC} = 3\sqrt{2}$
$\mathrm{BC} = 3\sqrt{2} \div \dfrac{\sqrt{3}}{2}$
$= 3\sqrt{2} \times \dfrac{2}{\sqrt{3}} = \dfrac{6\sqrt{2}}{\sqrt{3}}$
$= \dfrac{6\sqrt{6}}{3} = 2\sqrt{6}$