Câu 25.
\(\varphi=\varphi_1+\varphi_2=\dfrac{\pi}{2}-\dfrac{\pi}{3}=\dfrac{\pi}{6}\)
Định lí sin: \(\dfrac{A}{sin\alpha}=\dfrac{A_2}{sin\beta}\Rightarrow\dfrac{9}{sin\dfrac{\pi}{6}}=\dfrac{A_2}{sin\beta}\Rightarrow A_2=18sin\beta\ge18\)
\(\Rightarrow A_{2max}=18cm\)
Áp dụng phương trình: \(A^2=A_1^2+A_2^2+2A_1\cdot A_2\cdot cos\left(\varphi_2-\varphi_1\right)\)
\(\Rightarrow9^2=A_1^2+18^2+2\cdot A_1\cdot18\cdot cos\dfrac{5\pi}{6}\Rightarrow A_1^2-18\sqrt{3}A_1+243=0\)
\(\Rightarrow A_1=9\sqrt{3}cm\)
Chọn A.
Đúng 1
Bình luận (0)
