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प्रश्न
The vertices of a triangle ABC are A (0, 0), B (2, −1) and C (9, 2). Find cos B.
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उत्तर
We know that
\[\cos B = \frac{a^2 + c^2 - b^2}{2ac}\], where a = BC, b = CA and c = AB are the lengths of the sides of ∆ ABC
Thus,
\[a = BC = \sqrt{\left( 2 - 9 \right)^2 + \left( - 1 - 2 \right)^2} = \sqrt{49 + 9} = \sqrt{58}\]
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