Question

The vertices of a triangle abc are a(0,0),b(2,-1)andc(9,2)find cos b

Answer 1

$cosb = \frac{bc}{ab} \\ cosb = \frac{ \sqrt{ {(x2 - x1) }^{2} } + {(y2 - y1)}^{2} }{ \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } } \\ cosb = \frac{ \sqrt{ {(9 - 2)}^{2} + {(2 + 1)}^{2} } }{ \sqrt{ {(2 - 0)}^{2} + {(0 + 1)}^{2} } } \\ cosb = \frac{ \sqrt{ {7}^{2} } + {3}^{2} }{ \sqrt{ {2}^{2} + {1}^{2} } } \\ cosb = \frac{ \sqrt{49 + 9} }{ \sqrt{4 + 1} } \\ cosb = \sqrt{ \frac{58}{5} } \\ cosb = \sqrt{11.6}$

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