Question

三角形ABC中，如果a=sin10度，b=sin50度，角C=70度，求三角形ABC的面积

Answer 1

200000×1.5%×0.9434+200000×2%×0.89+200000×2.5%×0.8396+200000×0.8396
=200000×(1.5%×0.9434+2%×0.89+2.5%×0.8396+0.8396)
=200000×(1.5%×0.9434+2%×0.89+2.5%×0.8396+0.8396)
=200000×0.892541
=178508.2

Answer 2

S=1/2absinC
=1/2sin10sin50sin70
=sin10cos40cos20/2
=2sin10cos10cos20cos40/(4cos10)
=sin20cos20cos40/(4cos10)
=2sin20cos20cos40/(8cos10)
=sin40cos40/(8cos10)
=2sin40cos40/(16cos10)
=sin80/(16cos10)
=cos10/(16cos10)
=1/16