In a triangle ABC,b = sqrt{3},c = 1 and angle | vedclass

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(D) Given: $b = \sqrt{3}$,$c = 1$,and $\angle A = 30^\circ$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Substituting the values

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Given: $b = \sqrt{3}$,$c = 1$,and $\angle A = 30^\circ$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Substituting the values: $\cos 30^\circ = \frac{(\sqrt{3})^2 + 1^2 - a^2}{2(\sqrt{3})(1)}$. $\frac{\sqrt{3}}{2} = \frac{3 + 1 - a^2}{2\sqrt{3}}$. $3 = 4 - a^2$ $\Rightarrow a^2 = 1$ $\Rightarrow a = 1$. Since $b = \sqrt{3} \approx 1.732$ and $a = c = 1$,the side $b$ is the largest side. The largest angle is opposite the largest side,which is $\angle B$. Using the Law of Cosines for $\angle B$: $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{1^2 + 1^2 - (\sqrt{3})^2}{2(1)(1)} = \frac{1 + 1 - 3}{2} = -\frac{1}{2}$. Since $\cos B = -\frac{1}{2}$,we have $B = 120^\circ$.

In a triangle $ABC$,$b = \sqrt{3}$,$c = 1$ and $\angle A = 30^\circ$,then the largest angle of the triangle is ....$^\circ$

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