직각삼각형 삼각비 보기

[출제의도] 삼각비와 피타고라스 정리를 이용하여 삼각형에서 성립하는 내용을 추측한다.

직각삼각형 $\mathrm{ABC}$에서 피타고라스 정리에 의해 $\displaystyle \mathrm{\overline{BC}} ^{2} = \mathrm{\overline{AB}} ^{2} + \mathrm{\overline{AC}} ^{2}$이므로 $\displaystyle \mathrm{\overline{BC}} ^{2} = 3 ^{2} + 4 ^{2} = 25$, $\displaystyle \mathrm{\overline{BC}} = 5$ ㄱ. 삼각형 $\mathrm{ABC}$에서 $\mathrm{\triangle} ABC$$\displaystyle = \frac{1}{2} \times \mathrm{\overline{AB}} \times \overline{AC}$$\displaystyle = \frac{1}{2} \times \mathrm{\overline{BC}} \times \overline{AH}$ $\displaystyle \frac{1}{2} \times 3 \times 4 = \frac{1}{2} \times 5 \times \mathrm{\overline{AH}}$ $\displaystyle \mathrm{\overline{{AH}}} = \frac{12}{5}$ (참) ㄴ. $\displaystyle \tan \left( \mathrm{\angle} ADH \right) = \frac{\overline{AH}}{\overline{DH}} = 2$이므로 $\displaystyle \mathrm{\overline{DH}} = \frac{1}{2} \times \mathrm{\overline{AH}}$ ㄱ에서 $\displaystyle \mathrm{\overline{AH}} = \frac{12}{5}$이므로 $\displaystyle \mathrm{\overline{DH}} = \frac{1}{2} \times \mathrm{\overline{AH}}$$\displaystyle = \frac{1}{2} \times \frac{12}{5}$$\displaystyle = \frac{6}{5}$ 직각삼각형 $\mathrm{ABH}$에서 피타고라스 정리에 의해 $\displaystyle \mathrm{\overline{AB}} ^{2} = \mathrm{\overline{BH}} ^{2} + \mathrm{\overline{AH}} ^{2}$이므로 $\displaystyle \mathrm{\overline{BH}} ^{2} = \mathrm{9} - \frac{144}{25}$$\displaystyle = \frac{81}{25}$$\displaystyle = \left( \frac{9}{5} \right) ^{2}$ $\displaystyle \mathrm{\overline{BH}} = \frac{9}{5}$ 따라서 $\displaystyle \mathrm{\overline{BD}} = {\overline{BH}} + {\overline{DH}}$$\displaystyle = \frac{9}{5} + \frac{6}{5}$$= 3$ (거짓) ㄷ. $\displaystyle \mathrm{\overline{AB}} = \overline{BD} = 3$에서 삼각형 $\mathrm{ABD}$는 이등변삼각형이므로 $\mathrm{\angle} BAD = \mathrm{\angle} ADH$ 따라서 $\tan \left( \mathrm{\angle} BAD \right)$$= \tan \left( \mathrm{\angle} ADH \right)$$= 2$ (참) 이상에서 옳은 것은 ㄱ, ㄷ이다. [다른 풀이] ㄴ. $\displaystyle \tan \left( \mathrm{\angle} ADH \right) = \frac{\overline{AH}}{\overline{DH}} = 2$이므로 $\displaystyle \mathrm{\overline{DH}} = \frac{1}{2} \times \mathrm{\overline{AH}}$ ㄱ에서 $\displaystyle \mathrm{\overline{AH}} = \frac{12}{5}$이므로 $\displaystyle \mathrm{\overline{DH}}$$\displaystyle \mathrm{=} \frac{1}{2} \times \mathrm{\overline{AH}}$$\displaystyle = \frac{1}{2} \times \frac{12}{5}$$\displaystyle = \frac{6}{5}$ 두 삼각형 $\mathrm{ABH}$, $\mathrm{CBA}$에서 $\angle \mathrm{ABH}$는 공통인 각, $\mathrm{\angle} AHB = \angle CAB = 90 ^\circ$ 따라서 $\mathrm{\triangle} ABH \backsim \triangle CBA$이므로 $\displaystyle \mathrm{\overline{AB}} : \overline{BH} = \overline{CB} : \overline{BA}$ $\displaystyle \mathrm{3} : {\overline{BH}} = 5 : 3$ $\displaystyle 5 \times \mathrm{\overline{BH}} = 9$, $\displaystyle \mathrm{\overline{BH}} = \frac{9}{5}$ $\displaystyle \mathrm{\overline{BD}} = {\overline{BH}} + {\overline{DH}}$$\displaystyle = \frac{9}{5} + \frac{6}{5}$$= 3$ (거짓)