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Question Number 44365 by ajfour last updated on 27/Sep/18

Commented by ajfour last updated on 27/Sep/18

$F i n d a r e a o f △ A B C i n t e r m s o f$ $α, β, a n d R .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Sep/18

	$a r e a = \sqrt{s (s - a) (s - b) (s - c)}$ $\frac{a}{s i n α} = \frac{b}{s i n β} = \frac{c}{s i n {π - (α + β)}} = 2 R$ $a = 2 R s i n α = 4 R s i n \frac{α}{2} c o s \frac{α}{2}$ $b = 2 R s i n β = 4 R s i n \frac{β}{2} c o s \frac{β}{2}$ $c = 2 R s i n (α + β) = 4 R s i n \frac{α + β}{2} c o s \frac{α + β}{2}$ $s = \frac{a + b + c}{2} = R {(s i n α + s i n β + s i n (α + β)}$ $s = R {(2 s i n \frac{α + β}{2} c o s \frac{α - β}{2} + 2 s i n \frac{α + β}{2} c o s \frac{α + β}{2})}$ $s = 2 R s i n \frac{α + β}{2} \times 2 c o s \frac{α}{2} . c o s \frac{β}{2}$ $s = 4 R s i n \frac{α + β}{2} c o s \frac{α}{2} c o s \frac{β}{2}$ $s - a = 4 R c o s \frac{α}{2} (s i n \frac{α + β}{2} c o s \frac{β}{2} - s i n \frac{α}{2})$ $s - b = 4 R c o s \frac{β}{2} (s i n \frac{α + β}{2} c o s \frac{α}{2} - s i n \frac{β}{2})$ $s - c = 4 R s i n \frac{α + β}{2} (c o s \frac{α}{2} c o s \frac{β}{2} - c o s \frac{α + β}{2})$ $n o w p u t t i n g t h e v a l u e o f s, a, b a n d c$ $i n \sqrt{s (s - a) (s - b) (s - c)} i t i s s e e n t h a t a r e a o f$ $t r i a n g l e i s f (R, α, β)$
	Commented by ajfour last updated on 28/Sep/18

	$t h a n k s, a n y w a y S i r .$
	Answered by MrW3 last updated on 28/Sep/18

	$l e t O = c e n t e r o f c i r c l e$ $A C = 2 \times R \times \sin \frac{∠ A O C}{2} = 2 R \sin β$ $A B = 2 \times R \times \sin \frac{∠ A O B}{2} = 2 R \sin ∠ C = 2 R \sin (α + β)$ $A = A r e a o f Δ A B C = \frac{A B \times A C \times \sin α}{2}$ $\Rightarrow A = 2 R^{2} \sin α \sin β \sin (α + β)$
	Commented by ajfour last updated on 28/Sep/18

	$T h a n k y o u S i r, i m p o r t a n t r e s u l t!$ $\Rightarrow A = 2 R^{2} (a b c) {(\frac{1}{2 R})}^{3}$ $\Rightarrow R = \frac{a b c}{4 A} .$
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