In Delta ABC,if a = 2,b = 4,and angle C = 60^ | vedclass

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(C) Given: $a = 2$,$b = 4$,and $\angle C = 60^\circ$. Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. $\cos 60^\circ = \frac{2^2 + 4

Vedclass · Accepted Answer

$30^\circ, 90^\circ$

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(C) Given: $a = 2$,$b = 4$,and $\angle C = 60^\circ$. Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. $\cos 60^\circ = \frac{2^2 + 4^2 - c^2}{2(2)(4)} \Rightarrow \frac{1}{2} = \frac{4 + 16 - c^2}{16}$. $8 = 20 - c^2$ $\Rightarrow c^2 = 12$ $\Rightarrow c = 2\sqrt{3}$. Using the Sine Rule: $\frac{a}{\sin A} = \frac{c}{\sin C}$ $\Rightarrow \sin A = \frac{a \sin C}{c} = \frac{2 \sin 60^\circ}{2\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$. Thus,$\angle A = 30^\circ$. Since $\angle A + \angle B + \angle C = 180^\circ$,we have $30^\circ + \angle B + 60^\circ = 180^\circ$. Therefore,$\angle B = 90^\circ$.

In $\Delta ABC$,if $a = 2$,$b = 4$,and $\angle C = 60^\circ$,then $\angle A$ and $\angle B$ are equal to:

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