$The vertices of a triangle ABC are A (0, 0), B (2, - 1) a n d C (9, 2) . F i n d c o s B .$

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Solution

$W e k n o w t h a t c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} W h e r e a = B C, b = C A a n d C = A B a r e t h e s i d e s o f t h e t r i a n g l e A B C . W e h a v e, a = B C = \sqrt{(9 - 2)^{2} + (2 + 1)^{2}} = \sqrt{49 + 9} = \sqrt{5} 8 b = C A = \sqrt{(0 - 9)^{2} + (0 - 2)^{2}} = \sqrt{81 + 4} = \sqrt{8} 5 a n d, c = A B = \sqrt{(2 - 0)^{2} + (- 1 - 0)^{2}} = \sqrt{4 + 1} = \sqrt{5} ∴ c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{58 + 5 - 85}{2 \times \sqrt{5} 8 \times \sqrt{5}} = \frac{63 - 85}{2 \sqrt{290}} = \frac{- 22}{2 \sqrt{290}} = \frac{- 11}{\sqrt{290}} H e n c e, c o s B = \frac{- 11}{\sqrt{290}}$

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