How do you find the exact value of the third side given triangle ABC, a=6, b=4, mangleC=(2pi)/3?

Nov 19, 2017

$c = 2 \sqrt{19}$

Explanation:

Labelling triangle as diagram:

Using the Cosine Rule:

${c}^{2} = {a}^{2} + {b}^{2} - 2 a c \cdot \cos \left(C\right)$

${c}^{2} = {\left(6\right)}^{2} + {\left(4\right)}^{2} - \left(2\right) \left(6\right) \left(4\right) \cdot \cos \left(\frac{2 \pi}{3}\right)$

${c}^{2} = {\left(6\right)}^{2} + {\left(4\right)}^{2} - \left(2\right) \left(6\right) \left(4\right) \cdot \left(- \frac{1}{2}\right)$

${c}^{2} = 52 - \left(- 24\right) = 76$

${c}^{2} = 76 \implies c = \sqrt{76} = 2 \sqrt{19}$