∠ B C D = α \angle BCD=\alpha ∠ B C D = α \angle DAB=\beta$$(\dfrac{\pi}{2}<\beta<\pi) , A B ‾ = a \overline{AB}=a A B = a A D ‾ = b \overline{AD}=b A D = b a > b a>b a > b
삼각형 B C D BCD B C D
B D ‾ 2 = 3 2 + 2 2 − 2 ⋅ 3 ⋅ 2 ⋅ ( − 1 3 ) = 17
\overline{BD}^{2}
=3^{2}+2^{2}-2\cdot 3\cdot 2\cdot\left(-\dfrac{1}{3}\right)
=17
B D 2 = 3 2 + 2 2 − 2 ⋅ 3 ⋅ 2 ⋅ ( − 3 1 ) = 17 삼각형 A B D ABD A B D
a 2 + b 2 − 2 a b cos β = 17 ⋯ ⋯ ㉠
a^{2}+b^{2}-2ab\cos\beta=17\qquad\cdots\cdots\text{㉠}
a 2 + b 2 − 2 ab cos β = 17 ⋯⋯ ㉠ E E E A C AC A C 1 : 2 1:2 1 : 2 A E ‾ : C E ‾ = 1 : 2 \overline{AE}:\overline{CE}=1:2 A E : C E = 1 : 2
원의 지름이 A E ‾ \overline{AE} A E A P 1 P 2 AP_{1}P_{2} A P 1 P 2 r r r C E ‾ \overline{CE} C E C Q 1 Q 2 CQ_{1}Q_{2} C Q 1 Q 2 2 r 2r 2 r C E ‾ = 2 A E ‾ \overline{CE}=2\overline{AE} C E = 2 A E
사인법칙에 의해 아래와 같아요.
P 1 P 2 ‾ sin β = 2 r , Q 1 Q 2 ‾ sin α = 4 r
\dfrac{\overline{P_{1}P_{2}}}{\sin\beta}=2r,\qquad
\dfrac{\overline{Q_{1}Q_{2}}}{\sin\alpha}=4r
sin β P 1 P 2 = 2 r , sin α Q 1 Q 2 = 4 r P 1 P 2 ‾ : Q 1 Q 2 ‾ = 3 : 5 2 \overline{P_{1}P_{2}}:\overline{Q_{1}Q_{2}}=3:5\sqrt{2} P 1 P 2 : Q 1 Q 2 = 3 : 5 2
sin β sin α = P 1 P 2 ‾ 2 r ÷ Q 1 Q 2 ‾ 4 r = 2 P 1 P 2 ‾ Q 1 Q 2 ‾ = 6 5 2
\dfrac{\sin\beta}{\sin\alpha}
=\dfrac{\overline{P_{1}P_{2}}}{2r}\div\dfrac{\overline{Q_{1}Q_{2}}}{4r}
=\dfrac{2\overline{P_{1}P_{2}}}{\overline{Q_{1}Q_{2}}}
=\dfrac{6}{5\sqrt{2}}
sin α sin β = 2 r P 1 P 2 ÷ 4 r Q 1 Q 2 = Q 1 Q 2 2 P 1 P 2 = 5 2 6 cos α = − 1 3 \cos\alpha=-\dfrac{1}{3} cos α = − 3 1
sin α = 1 − cos 2 α = 2 2 3
\sin\alpha=\sqrt{1-\cos^{2}\alpha}=\dfrac{2\sqrt{2}}{3}
sin α = 1 − cos 2 α = 3 2 2 따라서 아래와 같아요.
sin β = 6 5 2 ⋅ 2 2 3 = 4 5
\sin\beta=\dfrac{6}{5\sqrt{2}}\cdot\dfrac{2\sqrt{2}}{3}=\dfrac{4}{5}
sin β = 5 2 6 ⋅ 3 2 2 = 5 4 π 2 < β < π \dfrac{\pi}{2}<\beta<\pi 2 π < β < π cos β < 0 \cos\beta<0 cos β < 0
cos β = − 1 − sin 2 β = − 3 5
\cos\beta=-\sqrt{1-\sin^{2}\beta}=-\dfrac{3}{5}
cos β = − 1 − sin 2 β = − 5 3 삼각형 A B D ABD A B D 2 2 2
1 2 a b sin β = 2 , 1 2 a b ⋅ 4 5 = 2 , a b = 5
\dfrac{1}{2}ab\sin\beta=2,\qquad
\dfrac{1}{2}ab\cdot\dfrac{4}{5}=2,\qquad ab=5
2 1 ab sin β = 2 , 2 1 ab ⋅ 5 4 = 2 , ab = 5 ㉠에 대입하면 아래와 같아요.
a 2 + b 2 − 2 ⋅ 5 ⋅ ( − 3 5 ) = 17
a^{2}+b^{2}-2\cdot 5\cdot\left(-\dfrac{3}{5}\right)=17
a 2 + b 2 − 2 ⋅ 5 ⋅ ( − 5 3 ) = 17 a 2 + b 2 = 11
a^{2}+b^{2}=11
a 2 + b 2 = 11 따라서 아래와 같아요.
( a + b ) 2 = a 2 + b 2 + 2 a b = 11 + 10 = 21
(a+b)^{2}=a^{2}+b^{2}+2ab=11+10=21
( a + b ) 2 = a 2 + b 2 + 2 ab = 11 + 10 = 21 a + b = 21
a+b=\sqrt{21}
a + b = 21 따라서 정답은 ① 예요.
