Ta có: \(\varphi_2 = 1,57 \approx \frac{\pi}{2} ; \varphi_1 = 0,35 rad \approx 20^0 \)
Vẽ giản đồ Fre-nen
Áp dụng định lý hàm số Sin ta có:
\(\triangle OMN: \)\(\frac{A_1}{\sin ONM} = \frac{A}{\sin OMN}\)<=> \(\frac{A_1}{\sin(\frac{\pi}{2} - \varphi)} = \frac{A}{\sin (\frac{\pi}{2} - \varphi_1)}\)=> \(A_1 = \frac{A}{\sin(\frac{\pi}{2}-\varphi_1)} \sin(\frac{\pi}{2} - \varphi).(1)\)
Tương tự: \(\triangle OHN: \) \(\frac{A_2}{\sin(\varphi_1+ \varphi)} = \frac{A}{\sin (\frac{\pi}{2} - \varphi_1)}\) => \(A_2 = \frac{A}{\sin(\frac{\pi}{2}-\varphi_1)} \sin(\varphi_1 + \varphi).(2)\)
Cộng (1) và (2) => \(A_1+A_2 = \frac{A}{\sin (\frac{\pi}{2} - \varphi_1)} (\sin (\frac{\pi}{2}-\varphi)+\sin(\varphi_1+ \varphi))\)
\( = \frac{A}{\sin (\frac{\pi}{2} - \varphi_1)} 2.\sin (\frac{\pi}{4}+\frac{\varphi_1}{2}).\cos(\frac{\pi}{4}-\frac{\varphi_1}{2}- \varphi)\) do áp dụng: \(\sin x + \cos x = 2 \sin (\frac{x+y}{2})\cos (\frac{x-y}{2}).\)
=> \((A_1+A_2)_{max} <=> \cos(\frac{\pi}{4}-\frac{\varphi_1}{2}- \varphi) =1\)
=> \((A_1+A_2)_{max} = \frac{A}{\sin (\frac{\pi}{2} - \varphi_1)} 2.\sin (\frac{\pi}{4}+\frac{\varphi_1}{2}) \)
\(= \frac{20}{\sin (90 - 20)} 2.\sin (45+\frac{20}{2}) \approx 34,88 cm.\)
Chọn đáp án.D.35cm.