# A triangle has sides A, B, and C. Sides A and B are of lengths 5 and 4, respectively, and the angle between A and B is (7pi)/12 . What is the length of side C?

Dec 31, 2015

#### Answer:

$c \cong 7.166$

#### Explanation:

By the Law of Cosines: ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(\Theta\right)$
In this example $a = 5 , b = 4 , \Theta = \frac{7 \pi}{12}$

Hence ${c}^{2} = {5}^{2} + {4}^{2} - 2. 5. 4 \cos \left(\frac{7 \pi}{12}\right)$
${c}^{2} = 25 + 16 - 40 \cos \left(\frac{7 \pi}{12}\right)$

Using a calculator $\cos \left(\frac{7 \pi}{12}\right) \cong - 0.2588190451$

Therefore ${c}^{2} \cong 41 + 40 \cdot 0.2588190451$

${c}^{2} \cong 51.3527616$
$c \cong \pm \sqrt{51.3527616}$

Since c must be positive
$c = 7.166$ To 3 decimal places