# A triangle has sides A, B, and C. Sides A and B are of lengths 3 and 14, respectively, and the angle between A and B is pi/6. What is the length of side C?

Jan 14, 2016

c ≅11.5

#### Explanation:

Law of cosines: ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \Theta$
Where a and b are sides opposite angles A and B and $\Theta$ is the included angle.

In this example $a = 3 , b = 14 \mathmr{and} \Theta = \frac{\pi}{6}$

Therefore: ${c}^{2} = {3}^{2} + {14}^{2} - 2. 3.14 \cos \left(\frac{\pi}{6}\right)$
${c}^{2} \cong 9 + 196 - 84 \cdot 0.8660254038$
$c \cong \sqrt{132.25386608}$
$c \cong \pm 11.5$ But c must be +ve

Therefore $c \cong 11.5$