The vertices of a triangle are $A (1, 1), B (4, 5)$ and $C (6, 13)$ . Find $cos A$

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Solution

Using law of cosines, we have, $a^{2} = b^{2} + c^{2} - 2 b c c o s A$
From the given points, $a^{2} = B C^{2} = 68$ , $b^{2} = A C^{2} = 169$ , $c^{2} = A B^{2} = 25$
Substituting these values, we have $c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$c o s A = \frac{169 + 25 - 68}{2 \times 13 \times 5}$ = $\frac{126}{130}$
$= \frac{63}{65}$

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