nicekikah

2020-12-13

Find the measure of each angle of triangle ABC, given AC = 6, $CB=2\surd 3$, and $AB=4\surd 3.$

Ayesha Gomez

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

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

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

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

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

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

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

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

$\angle A={\mathrm{cos}}^{-}1×\left(\sqrt{\frac{3}{2}}\right)$

$\angle A=30°$

Find $\angle B$

$AC²=CB²+AB²-2\left(CB\right)\left(AB\right)\mathrm{cos}\angle B:$

$2\left(CB\right)\left(AB\right)\mathrm{cos}\angle B=CB²+AB²-AC²$

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

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

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

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

$\angle B={\mathrm{cos}}^{-}1×\left(\frac{1}{2}\right)$

$\angle B=60°$

Find $\angle C$ using Triangle Sum Theorem:

$\angle A+\angle B+\angle C=180°$

$30°+60°+\angle C=180°$

$\angle C=90°$

Jeffrey Jordon