solution-to-Q212752-

Question Number 212867 by mr W last updated on 25/Oct/24

s o l u t i o n t o Q 212752

Commented by mr W last updated on 26/Oct/24

Commented by mr W last updated on 28/Oct/24

a s s u m e a ⩾ b

{A F}^{2} = l^{2} + c^{2} - 2 c l \cos θ

{A F}^{2} = a^{2} + b^{2} - 2 a b \cos (α + β)

\cos (α + β) = \frac{a^{2} + b^{2} - l^{2} - c^{2} + 2 l c \cos θ}{2 a b}

\Rightarrow \sin (α + β) = \sqrt{1 - {(\frac{a^{2} + b^{2} - l^{2} - c^{2} + 2 l c \cos θ}{2 a b})}^{2}}

O C = h

\frac{\sin α}{0.5 l} = \frac{\sin (90 ° - α - θ)}{h} = \frac{\cos (α + θ)}{h}

\frac{\sin β}{0.5 l} = \frac{\sin (90 ° - β + θ)}{h} = \frac{\cos (β - θ)}{h}

h = \frac{l \cos (α + θ)}{2 \sin α} = \frac{l \cos (β - θ)}{2 \sin β}

\frac{\sin α}{\sin β} = \frac{\cos (α + θ)}{\cos (β - θ)} = \frac{\cos α \cos θ - \sin α \sin θ}{\cos β \cos θ + \sin β \sin θ}

\Rightarrow \frac{1}{\tan α} - \frac{1}{\tan β} = 2 \tan θ

\frac{\sin φ}{c} = \frac{\sin θ}{A F}

\sin φ = \frac{c \sin θ}{\sqrt{l^{2} + c^{2} - 2 c l \cos θ}}

\Rightarrow φ = \sin^{- 1} \frac{c \sin θ}{\sqrt{l^{2} + c^{2} - 2 c l \cos θ}}

\frac{\sin (90 ° - β + φ + θ)}{a} = \frac{\sin (90 ° - α - φ - θ)}{b} = \frac{\sin (α + β)}{A F}

\frac{\cos (β - φ - θ)}{a} = \frac{\cos (α + φ + θ)}{b} = \frac{1}{λ}

w i t h λ = \frac{A F}{\sin (α + β)} = \sqrt{\frac{l^{2} + c^{2} - 2 c l \cos θ}{1 - {(\frac{a^{2} + b^{2} - l^{2} - c^{2} + 2 l c \cos θ}{2 a b})}^{2}}}

\cos (α + φ + θ) = \frac{b}{λ}

\Rightarrow α = \cos^{- 1} \frac{b}{λ} - φ - θ

\cos (β - φ - θ) = \frac{a}{λ}

\Rightarrow β = \cos^{- 1} \frac{a}{λ} + φ + θ

\Rightarrow \frac{1}{\tan (\cos^{- 1} \frac{b}{λ} - φ - θ)} - \frac{1}{\tan (\cos^{- 1} \frac{a}{λ} + φ + θ)} = 2 \tan θ

w e c a n s o l v e t h i s e q u a t i o n t o g e t θ .

\frac{T_{a}}{\sin β} = \frac{T_{b}}{\sin α} = \frac{m g}{\sin (α + β)}

\Rightarrow T_{a} = \frac{m g \sin β}{\sin (α + β)}

\Rightarrow T_{b} = \frac{m g \sin α}{\sin (α + β)}

Commented by mr W last updated on 28/Oct/24

Commented by mr W last updated on 28/Oct/24

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