In triangle ABC, a=12, b=8, c=8, how do you find the cosine of each of the angles?

Mar 28, 2017

Use the law of cosines as below

Explanation:

The law of cosines states that ${\gamma}^{2} = {\alpha}^{2} + {\beta}^{2} - 2 \alpha \beta \cos \Gamma$ where $\alpha$, $\beta$, and $\gamma$ are the sides of a triangle and $\Gamma$ is the angle opposite side $\gamma$ (the angle between sides $\alpha$ and $\beta$.

This law can be rearranged to solve for the cosine of an angle instead of a side.
$\cos \Gamma = \frac{{\alpha}^{2} + {\beta}^{2} - {\gamma}^{2}}{2 \alpha \beta}$

For each angle, plug in the adjacent sides for $\alpha$ and $\beta$ and the opposite side for $\gamma$

$\cos C = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b} = \frac{{12}^{2} + {8}^{2} - {8}^{2}}{2 \cdot 12 \cdot 8} = \frac{3}{4}$
$\cos B = \frac{{a}^{2} + {c}^{2} - {b}^{2}}{2 a c} = \frac{{12}^{2} + {8}^{2} - {8}^{2}}{2 \cdot 12 \cdot 8} = \frac{3}{4}$
$\cos A = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c} = \frac{{8}^{2} + {8}^{2} - {12}^{2}}{2 \cdot 8 \cdot 8} = \frac{\text{-} 1}{8}$