Question-143443

Question Number 143443 by mnjuly1970 last updated on 14/Jun/21

Commented by amin96 last updated on 14/Jun/21

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Answered by mr W last updated on 14/Jun/21

s a y R = r a d i u s

\cos α = \frac{R}{b}

{(b \sin α + a \cos α)}^{2} + {(a \sin α)}^{2} = R^{2}

b^{2} \sin^{2} α + 2 a b \sin α \cos α + a^{2} = b^{2} \cos^{2} α

a^{2} = b (b \cos 2 α - a \sin 2 α)

a^{2} = b \sqrt{a^{2} + b^{2}} (\cos φ \cos 2 α - \sin φ \sin 2 α)

\frac{a^{2}}{b \sqrt{a^{2} + b^{2}}} = \cos (φ + 2 α)

2 α = \cos^{- 1} \frac{a^{2}}{b \sqrt{a^{2} + b^{2}}} - \cos^{- 1} \frac{b^{2}}{b \sqrt{a^{2} + b^{2}}}

\cos 2 α = \frac{a (a b + \sqrt{a^{2} b^{2} + b^{4} - a^{4}})}{b (a^{2} + b^{2})}

\Rightarrow \tan α = \sqrt{\frac{b^{3} - a \sqrt{a^{2} b^{2} + b^{4} - a^{4}}}{2 a^{2} b + b^{3} + a \sqrt{a^{2} b^{2} + b^{4} - a^{4}}}}

Commented by mnjuly1970 last updated on 14/Jun/21

t h a n k s a l o t M r W \dots .