Question

数学几何求解

数学几何求解

Answer 1

不妨设CA=CB=1
则BA=BD=2sin20°
AD=4sin²20°=2-2cos40°
CD=BE=1-(2-2cos40°)=2(cos40°-cos60°)=4sin50°sin10°
ΔBDE中,∠DBE=30°
设∠BDE=60°-θ,∠BED=90°+θ, 且-50°<θ<60°
sin∠BED/sin∠BDE=BD/BE
sin(90°+θ)/sin(60°-θ)=(2sin20°)/(4sin50°sin10°)
cosθ/sin(60°-θ)=(sin20°cos10°)/(2sin50°sin10°cos10°)
cosθ/sin(60°-θ)=cos10°/sin50°
2sin50°cosθ=2sin(60°-θ)cos10°
sin(50°+θ)+sin(50°-θ)=sin(70°-θ)+sin(50°-θ)
sin(50°+θ)-sin(70°-θ)=0
2cos60°sin(θ-10°)=0
sin(θ-10°)=0,-50°<θ<60°
θ-10°=0°
θ=10°
所以∠BDE=60°-10°=50°

Answer 2

算出来了可以和我说说吗？