已知A(0,a),B(0,b)是Y轴正正半轴上两点0,
已知A(0,a),B(0,b)是Y轴正正半轴上两点0, 已知A(0,a),B(0,b)是Y轴正正半轴上两点0, C点为（a,0)时 设直线AC和BC的倾斜角分别为α, β. 全部展开 设直线AC和BC的倾斜角分别为α, β. 收起
已知A(0,a),B(0,b)是Y轴正正半轴上两点0,
设直线AC和BC的倾斜角分别为α,β.
tanα = (a-0)/(0-x) = -a/x
tanβ = (b-0)/(0-x) = -b/x
显然角ACB一定是锐角 = α-β
f(x)= tan(α- β) = (tanα - tanβ)/(1 + tanα*tanβ) = (-a/x + b/x)/(1 + ab/x²)= (b-a)x/(x²+ab)
f'(x) = (b-a)[1/(x²+ ab) + x*(-1)*2x/(x²+ab)²]
= (b-a)(ab - x²)/(x²+ab)² = 0
x = √(ab)
C(√(ab),0)时角ACB最大
tanα = (a-0)/(0-x) = -a/x
tanβ = (b-0)/(0-x) = -b/x
显然角ACB一定是锐角 = α-β
f(x)= tan(α- β) = (tanα - tanβ)/(1 + tanα*tanβ) = (-a/x + b/x)/(1 + ab/x²)= (b-a)x/(x...
tanα = (a-0)/(0-x) = -a/x
tanβ = (b-0)/(0-x) = -b/x
显然角ACB一定是锐角 = α-β
f(x)= tan(α- β) = (tanα - tanβ)/(1 + tanα*tanβ) = (-a/x + b/x)/(1 + ab/x²)= (b-a)x/(x²+ab)
f'(x) = (b-a)[1/(x²+ ab) + x*(-1)*2x/(x²+ab)²]
= (b-a)(ab - x²)/(x²+ab)² = 0
x = √(ab)
C(√(ab), 0)时角ACB最大