How do you find the exact value of the third side given $△ A B C$ $triangle ABC$ , a=6, b=4, $m ∠ C = \frac{2 π}{3}$ $mangleC=(2pi)/3$ ?

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Answer 1 · 2017-11-19T17:28:02

$c = 2 \sqrt{19}$ $c= 2sqrt(19)$

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Labelling triangle as diagram:

Using the Cosine Rule:

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

$c^{2} = {(6)}^{2} + {(4)}^{2} - (2) (6) (4) \cdot cos (\frac{2 π}{3})$ $c^2=(6)^2+(4)^2-(2)(6)(4)*cos((2pi)/3)$

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

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

$c^{2} = 76 \Rightarrow c = \sqrt{76} = 2 \sqrt{19}$ $c^2=76=>c=sqrt(76)=2sqrt(19)$

How do you find the exact value of the third side given triangle ABC△ABCtriangle ABC, a=6, b=4, mangleC=(2pi)/3m∠C=2π3mangleC=(2pi)/3?