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在三角形ABC中AB=AC,角ABD等于40度,BD平分角ABC,延长BD至E,使DE=AD,角ECA度数为多
人气:184 ℃ 时间:2020-02-05 19:19:26
解答

∠ABD=∠DBC=40°, ∠ABC=∠ACB=80°,∠BAC=20°,∠BDC=60°
∠ADE=∠ABD+∠BAD=60°,DE=AD,△ADE为等边三角形.
∠ADE=∠AED=∠EAD=60°
设AD=DE=AE=x,
则AB/sin∠AEB = AE/sin∠ABE
AB=x*sin60°/sin40°
BC/sin∠BAC = AB/sin∠ACB
BC=x*sin60°* sin 20°/ (sin40°* sin80°)
CD/sin∠CBD = BC/sin∠BDC
CD=x*sin60°* sin 20°*sin40°/ (sin40°* sin80°*sin60°)
=x*sin20°/sin80°=2xsin10°
DE/sin∠DCE = CD/ sin∠CED ∠DCE + ∠CED =∠ADE=60° 设∠ECA=a
即 x/(sina)=2xsin10°/sin(60°-a)
tana=sin60°/ (cos60°+2sin10°)
a=arctan(sin60°/ (cos60°+2sin10°))= 45.626299575261°
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