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In triangle ABC, if a=2, b=1 and \angle C=60^{\circ} then find other two angles ?

Option: 1

\frac{\pi}{6},\frac{\pi}{2}


Option: 2

\frac{\pi}{3},\frac{\pi}{3}


Option: 3

\frac{\pi}{12},\frac{7\pi}{12}


Option: 4

None of these


Answers (1)

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\\\cos C= \frac{a^2+b^2-c^2}{2ab}\\\\ \frac{1}{2}=\frac{4+1-c^2}{4}\\ c=\sqrt{3}\\ \\Also,\,\,\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \frac{\sin A}{2}=\frac{\sin B}{1}=\frac{\frac{\sqrt{3}}{2}}{\sqrt{3}}\\\sin A = 1\Rightarrow A = \frac{\pi}{2}\\ \sin B=\frac{1}{2}\Rightarrow B=\frac{\pi}{6}

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Divya Prakash Singh

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