等腰三角形ABC中AC=BC,且角ACB=106度。M是三角形內一點使得角MAC=7度及角MCA=23度,求角CMB。
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let C(0,sin37'), A(cos37',0),B(-cos37',0),M(x,y)
MA: y=-(x-cos37)tan30'
MC: y=sin37'-xtan60'
==> x=(sin37'cos30'-cos37'sin30')/cos30' /(tan60'-tan30')
=sin7'
y=sin37'-sin7'*sqrt(3)
=sin7cos30+cos7sin30-sin7cos30*2
=sin23'
x+cos37'=sin7'+cos7'cos30'-sin7'/2
=cos23'
slop of BM
=y/(x+cos37')=tan23'
slop of CM
=-tan60'
so, angle CMB=23'+60'=83'