求角的度数值？

one海洋

northwolves

正弦定理
$\frac{sin(\angleAFB)}{sin(\angleABF)}=\frac{AB}{AF}=\frac{AB}{AE}=\frac{sin(\angleE)}{sin(\angleABE)}$
$\frac{sin(\angleAFB)}{sin127.5°}=\frac{sin(\frac{180°-63.75°}{2})}{sin63.75°}$
$\angleAFB=arcsin(2cos(63.75°)cos(31.875°))$

one海洋

northwolves 发表于 2022-5-16 11:43
正弦定理
$\frac{sin(\angleAFB)}{sin(\angleABF)}=\frac{AB}{AF}=\frac{AB}{AE}=\frac{sin(\angleE)}{sin ...

几何画板软件，显示角的度数值误差好大，不相符

王守恩

one海洋 发表于 2022-5-16 11:58
几何画板软件，显示角的度数值误差好大，不相符

答案正确！谢谢northwolves!

解法一：
\(AB=1\ \ \ \ \ \ \frac{1}{\sin(58.125^\circ)}=\frac{AE}{\sin(63.75^\circ)}\)
\(AF=AE\ \ \ \ \frac{AF}{\sin(127.5^\circ)}=\frac{1}{\sin(∠AFB)}\)

解法二：
\(AE=2\ \ \ \ \ \ \ \ \frac{2}{\sin(63.75^\circ)}=\frac{AB}{\sin(58.125^\circ)}\)
\(AF=AE\ \ \ \ \frac{2}{\sin(127.5^\circ)}=\frac{AB}{\sin(∠AFB)}\)

解法三：
\(AE/2=1\ \ \ \ \ \frac{1}{\sin(31.875^\circ)}=\frac{AB}{\sin(90^\circ)}\)
\(AF=AE\ \ \ \ \ \ \frac{2}{\sin(127.5^\circ)}=\frac{AB}{\sin(∠AFB)}\)

解法四：角格点问题。记\(∠AFB=2a\)
\(\D1=\frac{\sin(29.0625^\circ-a)\sin(5.625^\circ+2a)\sin(63.75^\circ)\sin(2a)}{\ \ \sin(58.125^\circ)\sin(52.5^\circ-2a)\sin(63.75^\circ)\sin(87.1875^\circ-a)\ \ \ }\)

one海洋

one海洋 发表于 2022-5-16 11:58
几何画板软件，显示角的度数值误差好大，不相符

几何画板软件，显示出角的度数相符