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-2-2-a-2-then-find-angle-A-

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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