在 ΔABC 中，已知 ∠A=24°，∠C=30°，D 是 AC 上的一点，AB=CD ，求 ∠DBA

王守恩

\(网上有这么一道题:△ABC,∠A=24^\circ,∠C=30^\circ,AB=CD(D是AC上的点),求∠DBA\)
三角函数解题还是有优势的。给出部分解法，有兴趣的网友不妨一试。
\(01,设∠DBA=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(30^\circ)*\sin(30^\circ)*\sin(\theta)}{\sin(24^\circ)*(\sin(54^\circ)-\sin(30^\circ))*\sin(\theta+54^\circ)}=1,解得\theta=30^\circ\)
\(02,设∠BDC=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(30^\circ)*\sin(30^\circ)*\sin(\theta-24^\circ)}{\sin(24^\circ)*(\sin(54^\circ)-\sin(30^\circ))*\sin(\theta+30^\circ)}=1,解得\theta=54^\circ\)
\(03,设∠DBC=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(30^\circ)*\sin(30^\circ)*\sin(\theta+54^\circ)}{\sin(24^\circ)*(\sin(54^\circ)-\sin(30^\circ))*\sin(\theta)}=1,解得\theta=96^\circ\)
\(04,设∠BDA=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(30^\circ)*\sin(30^\circ)*\sin(\theta+24^\circ)}{\sin(24^\circ)*(\sin(54^\circ)-\sin(30^\circ))*\sin(\theta-30^\circ)}=1,解得\theta=126^\circ\)
\(05,设∠DBA=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(\theta+24^\circ)*\sin(\theta+24^\circ)*\sin(\theta)}{\sin(\theta+24^\circ)\sin(24^\circ)/\sin(30^\circ)*\sin(\theta)*\sin(\theta+54^\circ)}=1,解得\theta=30^\circ\)
\(06,设∠BDC=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(\theta)*\sin(\theta)*\sin(\theta-24^\circ)}{\sin(\theta)\sin(24^\circ)/\sin(30^\circ)*\sin(\theta-24^\circ)*\sin(\theta+30^\circ)}=1,解得\theta=54^\circ\)
\(07,设∠DBC=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(\theta+30^\circ)*\sin(\theta+30^\circ)*\sin(\theta+54^\circ)}{\sin(\theta+30^\circ)\sin(24^\circ)/\sin(30^\circ)*\sin(\theta+54^\circ)*\sin(\theta)}=1,解得\theta=96^\circ\)
\(08,设∠BDA=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(\theta)*\sin(\theta)*\sin(\theta+24^\circ)}{\sin(\theta)\sin(24^\circ)/\sin(30^\circ)*\sin(\theta+24^\circ)*\sin(\theta-30^\circ)}=1,解得\theta=126^\circ\)
\(09,设∠DBA=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(\theta+54^\circ)*\sin(\theta+54^\circ)*\sin(\theta)}{\sin(\theta+24^\circ)*\sin(\theta)\sin(30^\circ)/\sin(24^\circ)*\sin(\theta+54^\circ)}=1,解得\theta=30^\circ\)
\(10,设∠BDC=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(\theta+30^\circ)*\sin(\theta+30^\circ)*\sin(\theta-24^\circ)}{\sin(\theta)*\sin(\theta-24^\circ)\sin(30^\circ)/\sin(24^\circ)*\sin(\theta+30^\circ)}=1,解得\theta=54^\circ\)
\(11,设∠DBC=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(\theta)*\sin(\theta)*\sin(\theta+54^\circ)}{\sin(\theta+30^\circ)*\sin(\theta+54^\circ)\sin(30^\circ)/\sin(24^\circ)*\sin(\theta)}=1,解得\theta=96^\circ\)
\(12,设∠BDA=\theta,\frac{AB*CD*\sin(∠DBA)}{CB*AD*\sin(∠DBC)}=\frac{\sin(\theta-30^\circ)*\sin(\theta-30^\circ)*\sin(\theta+24^\circ)}{\sin(\theta)*\sin(\theta+24^\circ)\sin(30^\circ)/\sin(24^\circ)*\sin(\theta-30^\circ)}=1,解得\theta=126^\circ\)
\(......\)

天山草

给楼主的问题配上图：

天山草

不解三角方程行吗？ 直接算出  

\(θ =
FullSimplify[\frac{180}{π}
   arctan\frac{0.5(\sqrt{5} - 1)sin[24π /180]}{
    1 - 0.5 (\sqrt{5} - 1) cos[24 π /180]}=30°\)

王守恩

\(在△ABC中,已知\sin(A)+\sin(B)+\sin(C)=\frac{54}{25},求\sin(A)\sin(B)\sin(C)的最大值。\)

天山草

对于 4# 楼的问题，用拉格朗日乘数法做，结果是当  \( A=B=2 arctan(1/3) \) 时

\(sinA sinB sinC \)有最大值 \( 216/625 = 0.3456\)。   

当 \( A =B = 2 arctan[\frac{1}{9} (-1+ 4 \sqrt{14} sinh[\frac{1}{3} arcsinh[\frac{589}{56\sqrt{14}}]])]  \) 时

有最小值 \( 0.1658636882601592……\)。

王守恩

天山草 发表于 2022-1-28 15:38
对于 4# 楼的问题，用拉格朗日乘数法做，结果是当  \( A=B=2 arctan(1/3) \) 时

\(sinA sinB sinC \)有 ...

谢谢天山草！这一块我是盲区，谢谢天山草！能再来一个？

\(在△ABC中,已知\sin(A)+\sin(B)+\sin(C)=\frac{1}{6},求\sin(A)\sin(B)\sin(C)的最大值。\)

天山草

当三个角度的正弦和为 1/6 时，三个角度的正弦积的最大值为  0.00014473880 ... ...