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the vertices of triangle ABC are A(0,0) B(2,-1) C(9,2),find cosB

$A B^{2} = (2 - 0)^{2} + (- 1 - 0)^{2} = 5 B C^{2} = (2 - 9)^{2} + (- 1 - 2)^{2} = 58 A C^{2} = (0 - 9)^{2} + (0 - 2)^{2} = 85 \cos B = \frac{A B^{2} + B C^{2} - A C^{2}}{2 A B \times B C} = \frac{5 + 58 - 85}{2 \sqrt{5} \sqrt{58}} = - \frac{11}{\sqrt{290}}$

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