Question-97557

Question Number 97557 by Power last updated on 08/Jun/20

Answered by mr W last updated on 08/Jun/20

∠ C B E = ∠ E B D = θ = \frac{1}{2} (\frac{π}{2} - α)

\Rightarrow 2 θ = \frac{π}{2} - α

B D = 2 r \sin α

E B = \frac{2 r}{\cos θ}

E D = \sqrt{{E B}^{2} + {B D}^{2} - 2 \times E B \times B D \times \cos θ}

E D = \sqrt{\frac{4 r^{2}}{\cos^{2} θ} + 4 r^{2} \sin^{2} α - 8 r^{2} \sin α}

E D = 2 r \sqrt{\frac{1}{\cos^{2} θ} + \sin^{2} α - 2 \sin α}

\frac{\sin φ}{E B} = \frac{\sin θ}{E D}

\sin φ = \frac{E B \sin θ}{E D} = \frac{2 r \sin θ}{\cos θ 2 r \sqrt{\frac{1}{\cos^{2} θ} + \sin^{2} α - 2 \sin α}}

\sin φ = \frac{\sin θ}{\sqrt{1 + (\sin^{2} α - 2 \sin α) \cos^{2} θ}}

\sin φ = \frac{\sqrt{1 - \cos 2 θ}}{\sqrt{2 + (\sin^{2} α - 2 \sin α) (\cos 2 θ + 1)}}

\sin φ = \frac{\sqrt{1 - \sin α}}{\sqrt{2 - (2 - \sin α) (1 + \sin α) \sin α}}

\Rightarrow φ = \sin^{- 1} \frac{\sqrt{1 - \sin α}}{\sqrt{2 - (2 - \sin α) (1 + \sin α) \sin α}}

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