In triangle ABC with usual notations b=sqrt{3 | vedclass

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(D) By the cosine rule,we have: $a^2 = b^2 + c^2 - 2bc \cos A$ $a^2 = (\sqrt{3})^2 + (1)^2 - 2(\sqrt{3})(1) \cos(30^{\circ})$ $a^2 = 3 + 1 - 2\sqrt{3}

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) By the cosine rule,we have: $a^2 = b^2 + c^2 - 2bc \cos A$ $a^2 = (\sqrt{3})^2 + (1)^2 - 2(\sqrt{3})(1) \cos(30^{\circ})$ $a^2 = 3 + 1 - 2\sqrt{3} \left(\frac{\sqrt{3}}{2}ight)$ $a^2 = 4 - 3 = 1$ $	herefore a = 1$ Since $b = \sqrt{3} \approx 1.732$ is the largest side,the largest angle is $\angle B$. Using the cosine rule for $\angle B$: $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{1^2 + 1^2 - (\sqrt{3})^2}{2(1)(1)} = \frac{1 + 1 - 3}{2} = -\frac{1}{2}$ $	herefore B = 120^{\circ}$

In triangle $ABC$ with usual notations $b=\sqrt{3}$,$c=1$,and $m \angle A=30^{\circ}$,then the largest angle of the triangle is (in $^{\circ}$)

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