직사각형과 탄젠트

[출제의도] 이해능력-삼각함수

$\displaystyle {\overline{DP}} = 2 a ( a > 0 )$, $\displaystyle {\overline{PC}} = 3 a$라 하면 $\displaystyle {\overline{AB}} = 5 a$이고 $\displaystyle {\overline{AP}} = \sqrt{( 3 \sqrt{2} ) ^{2} + ( 2 a ) ^{2}} = \sqrt{18 + 4 a ^{2}}$ 삼각형 $ABQ$와 삼각형 $PAD$는 닮음이므로 $\displaystyle {\overline{AB}} : {\overline{BQ}} = {\overline{PA}} : {\overline{AD}}$ $\displaystyle 5 a : a = \sqrt{18 + 4 a ^{2}} : 3 \sqrt{2}$ $18 + 4 a ^{2} = 18 a ^{2}$ $\displaystyle \therefore a = \frac{3}{\sqrt{7}} = \frac{3 \sqrt{7}}{7}$ $\displaystyle \therefore {\overline{DP}} = \frac{6 \sqrt{7}}{7} , {\overline{PC}} = \frac{9 \sqrt{7}}{7}$ 점 $Q$에서 직선 $BC$에 내린 수선의 발을 $H$라 하고 $\angle BQH = \alpha$, $\angle CQH = \beta$ 라 하자. $\tan \alpha = \tan ( \angle ABQ ) = \tan ( \angle PAD )$ $\displaystyle = \frac{{\overline{DP}}}{{\overline{AD}}} = \frac{\frac{6 \sqrt{7}}{7}}{3 \sqrt{2}} = \frac{\sqrt{14}}{7}$ $\cdots$㉠ 한편, $\displaystyle {\overline{AB}} : {\overline{BQ}} = {\overline{BQ}} : {\overline{QH}}$에서 $\displaystyle {\overline{QH}} = \frac{{\overline{BQ}} ^{2}}{{\overline{AB}}} = \frac{5 \sqrt{7}}{3}$이므로 $\displaystyle {\overline{BH}} = {\overline{QH}} \tan \alpha = \frac{5 \sqrt{7}}{3} \times \frac{\sqrt{14}}{7} = \frac{5 \sqrt{2}}{3}$ 이다. $\displaystyle \therefore {\overline{HC}} = {\overline{BC}} - {\overline{BH}} = 3 \sqrt{2} - \frac{5 \sqrt{2}}{3} = \frac{4 \sqrt{2}}{3}$ $\displaystyle \therefore \tan \beta = \frac{{\overline{HC}}}{{\overline{QH}}} = \frac{\frac{4 \sqrt{2}}{3}}{\frac{5 \sqrt{7}}{3}} = \frac{4 \sqrt{2}}{5 \sqrt{7}} = \frac{4 \sqrt{14}}{35}$ $\displaystyle \therefore \tan ( \angle BQC ) = \tan ( \alpha + b ) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$ $\displaystyle = \frac{\frac{\sqrt{14}}{7} + \frac{4 \sqrt{14}}{35}}{1 - \frac{\sqrt{14}}{7} \times \frac{4 \sqrt{14}}{35}} = \frac{\frac{9 \sqrt{14}}{35}}{1 - \frac{8}{35}} = \frac{9 \sqrt{14}}{27} = \frac{\sqrt{14}}{3}$