Find the measure of each angle of triangle ABC, given AC = 6, CB=2√3, and...

Use the Law of Cosines to find two angle measures and the Triangle Sum Theorem to find the third angle.
Find ${\displaystyle \angle A:}$

$CB²=AC²+AB²-2(AC)(AB)\cos \angle A$

$2(AC)(AB)\cos \angle A=AC²+AB²-CB²$

$\cos \angle A=\frac{AC²+AB²-CB²}{2(AC)(AB)}$

$\cos \angle A=(6²+(4\sqrt{3})²-\frac{2\sqrt{3})²}{2(6)(4\sqrt{3})}$

$\cos \angle A=\frac{72}{48}\sqrt{3}$

$\cos \angle A=\frac{3}{2}\sqrt{3}$

$\cos \angle A=\sqrt{\frac{3}{2}}$

$\angle A={\cos }^{-}1\times (\sqrt{\frac{3}{2}})$

$\angle A=30°$

Find $\angle B$

$AC²=CB²+AB²-2(CB)(AB)\cos \angle B:$

$2(CB)(AB)\cos \angle B=CB²+AB²-AC²$

$\cos \angle B=\frac{CB²+AB²-AB²}{2(CB)(AB)}$

$\cos \angle B=\frac{(2\sqrt{3})²+(4\sqrt{3})²-6²}{2(2\sqrt{3})(4\sqrt{3})}$

$\cos \angle B=\frac{24}{48}$

$\cos \angle B=\frac{1}{2}$

$\angle B={\cos }^{-}1\times (\frac{1}{2})$

$\angle B=60°$

Find $\angle C$ using Triangle Sum Theorem:

$\angle A+\angle B+\angle C=180°$

$30°+60°+\angle C=180°$

$\angle C=90°$

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