Question-48763

Question Number 48763 by mr W last updated on 28/Nov/18

Commented by mr W last updated on 28/Nov/18

F i n d R i n t e r m s o f a, b, c .

Answered by ajfour last updated on 28/Nov/18

Answered by ajfour last updated on 28/Nov/18

R \sin θ = \frac{a}{2}

2 R \cos (θ + α) = \sqrt{b^{2} + c^{2}}

\Rightarrow 2 R (\cos θ \cos α - \sin θ \sin α) = \sqrt{b^{2} + c^{2}}

2 R (\cos θ \times \frac{b}{\sqrt{b^{2} + c^{2}}} - \frac{a}{2 R} \times \frac{c}{\sqrt{b^{2} + c^{2}}}) = \sqrt{b^{2} + c^{2}}

\Rightarrow 2 b R \cos θ = b^{2} + c^{2} + a c

o r (4 R^{2} - a^{2}) b^{2} = {(b^{2} + c^{2} + a c)}^{2}

4 R^{2} = a^{2} + \frac{{(b^{2} + c^{2} + a c)}^{2}}{b^{2}}

R = \frac{1}{2} \sqrt{a^{2} + \frac{{(b^{2} + c^{2} + a c)}^{2}}{b^{2}}} .

Commented by mr W last updated on 28/Nov/18

c o r r e c t! t h a n k y o u s i r!

Commented by mr W last updated on 28/Nov/18

a l t e r n a t i v e w a y :

c i r c u m c i r c l e o f a t r i a n g l e w i t h s i d e s :

a

\sqrt{b^{2} + c^{2}}

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

R = \frac{a \sqrt{(b^{2} + c^{2}) (b^{2} + {(a + c)}^{2})}}{4 A}

A = \frac{a b}{2}

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

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

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