# How do you use the law of cosines to find AB given a triangle ABC with Angle C=41, AC=13 and CB=29?

May 2, 2018

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$

$= {29}^{2} + {13}^{2} - 2 \left(29\right) \left(13\right) \cos {41}^{\circ}$$= 1010 - 754 \setminus \cos {41}^{\circ}$

$c \approx 21.00$

#### Explanation:

Typically we label the opposite sides with small letters, so we have angle $C = {41}^{\circ}$ and $b = 13$ and $a = 29$. The Law of Cosines is

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$$= {29}^{2} + {13}^{2} - 2 \left(29\right) \left(13\right) \cos {41}^{\circ}$$= 1010 - 754 \cos {41}^{\circ}$

$c \approx 21.00$