tanθ=xとおくと
x²-(1+√3)x+√3=0
x²-x-√3x+√3=0
x(x-1)-√3(x-1)=0
(x-√3)(x-1)=0
x=√3,1
よってθ=45°、60°
tanθ=xとおくと
x²-(1+√3)x+√3=0
x²-x-√3x+√3=0
x(x-1)-√3(x-1)=0
(x-√3)(x-1)=0
x=√3,1
よってθ=45°、60°
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なるほど!
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