If a,b, and A of a triangle are fixed and two possible values of the third side be c_1 and c_2 such that c_1 ^2 +c_1 c_2 +c


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Question Number 21889 by ajfour last updated on 06/Oct/17

$I f a, b, a n d A o f a t r i a n g l e a r e$ $f i x e d a n d t w o p o s s i b l e v a l u e s o f t h e$ $t h i r d s i d e b e c_{1} a n d c_{2} s u c h t h a t$ $c_{1}^{2} + c_{1} c_{2} + c_{2}^{2} = a^{2}, t h e n f i n d a n g l e A .$
	Answered by mrW1 last updated on 06/Oct/17

	$a^{2} = b^{2} + c^{2} - 2 bc \times \cos A$ $\Rightarrow c^{2} - (2 b \cos A) c + (b^{2} - a^{2}) = 0$ $c_{1} + c_{2} = 2 b \cos A$ $c_{1} c_{2} = b^{2} - a^{2}$ $c_{1}^{2} + c_{1} c_{2} + c_{2}^{2} = a^{2}$ $c_{1}^{2} + 2 c_{1} c_{2} + c_{2}^{2} = a^{2} + c_{1} c_{2}$ ${(c_{1} + c_{2})}^{2} = a^{2} + c_{1} c_{2}$ $\Rightarrow {4 b}^{2} \cos^{2} A = a^{2} + b^{2} - a^{2} = b^{2}$ $\Rightarrow 4 \cos^{2} A = 1$ $\Rightarrow \cos A = \pm \frac{1}{2}$ $\Rightarrow A = 60 ° or 120 °$
	Commented by ajfour last updated on 06/Oct/17

	$t h a n k y o u s i r .$
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