Application of differentiation

ai) By sine law,
x/sinθ = 5/sin(π/3 -θ)
x= 5sinθ/(sinπ/3cosθ - cosπ/3sinθ)
= 10sinθ/√3cosθ-sinθ

ii)
when x =5,

5= 10sinθ/√3cosθ-sinθ
3cosθ-sinθ=2sinθ
tanθ=1/√3
θ=π/6

b)x = 10sinθ/√3cosθ-sinθ
Diff. both sides w.r.t. t
dx/dt = dθ/dt x [10cosθ(√3cosθ-sinθ)-10sinθ(-√3sinθ-cosθ)] / (√3cosθ-sinθ)²
-0.5 = dθ/dt x 10[√3(cos²θ+sin²θ)-sinθcosθ+sinθcosθ] / (√3cosθ-sinθ)²
dθ/dt= -0.5(√3cosθ-sinθ)² / 10
= -0.05(√3cosθ-sinθ)² = -sin²(π/3 -θ)

When P is at a distance 5m from B, θ=π/6
dθ/dt | θ=π/6 = -sin²(π/3 -π/6) = -0.25


2011-03-06 09:34:18 補充：
sin(A+B) = sinAcosB+cosAsinB
sin(A-B) = sinAcosB - cosAsinB

呢個係compound angle formula

抱歉, 想再問你一樣野
請問x/sinθ = 5/sin(π/3 -θ)
怎樣換成
x= 5sinθ/(sinπ/3cosθ - cosπ/3sinθ)
= 10sinθ/√3cosθ-sinθ