Question-163736

Question Number 163736 by HongKing last updated on 09/Jan/22

Answered by mr W last updated on 10/Jan/22

Commented by mr W last updated on 10/Jan/22

R = \frac{a}{2 \sin A}

Δ = \frac{b c \sin A}{2}

r = \frac{2 Δ}{a + b + c} = \frac{b c \sin A}{a + b + c}

O I / / A B \Leftrightarrow R \cos A = r

\frac{a}{2 \sin A} \cos A = \frac{b c \sin A}{a + b + c}

\cos A = \frac{2 b c \sin^{2} A}{a (a + b + c)}

\cos A = \frac{2 b c (1 - \cos A) (1 + \cos A)}{a (a + b + c)}

\frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{2 b c (1 - \frac{b^{2} + c^{2} - a^{2}}{2 b c}) (1 + \frac{b^{2} + c^{2} - a^{2}}{2 b c})}{a (a + b + c)}

b^{2} + c^{2} - a^{2} = \frac{[a^{2} - {(b - c)}^{2}] [{(b + c)}^{2} - a^{2}]}{a (a + b + c)}

b^{2} + c^{2} - a^{2} = \frac{[a^{2} - {(b - c)}^{2}] (b + c - a)}{a}

(b + c) a^{2} - 2 b c a - (b^{2} - c^{2}) (b - c) = 0

a = \frac{b c + \sqrt{b^{2} c^{2} + (b + c) (b^{2} - c^{2}) (b - c)}}{b + c}

a = \frac{b c + \sqrt{b^{2} c^{2} + {(b^{2} - c^{2})}^{2}}}{b + c}

\Rightarrow a = \frac{b c + \sqrt{b^{4} + c^{4} - b^{2} c^{2}}}{b + c} ✓

Commented by Tawa11 last updated on 10/Jan/22

Great sir

Commented by HongKing last updated on 10/Jan/22

very nice my dear Sir thank you so much