In a triangle ABC,if a = 2,B = 60^circ and C | vedclass

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(B) Given: $a = 2$,$B = 60^\circ$,$C = 75^\circ$.In any triangle $ABC$,the sum of angles is $180^\circ$,so $A = 180^\circ - (B + C) = 180^\circ - (60^

Vedclass · Accepted Answer

$\sqrt{6}$

Vedclass · Answer

(B) Given: $a = 2$,$B = 60^\circ$,$C = 75^\circ$.In any triangle $ABC$,the sum of angles is $180^\circ$,so $A = 180^\circ - (B + C) = 180^\circ - (60^\circ + 75^\circ) = 180^\circ - 135^\circ = 45^\circ$.Using the Sine Rule: $\frac{b}{\sin B} = \frac{a}{\sin A}$.Substituting the values: $\frac{b}{\sin 60^\circ} = \frac{2}{\sin 45^\circ}$.$b = \frac{2 	imes \sin 60^\circ}{\sin 45^\circ} = \frac{2 	imes (\frac{\sqrt{3}}{2})}{\frac{1}{\sqrt{2}}} = \sqrt{3} 	imes \sqrt{2} = \sqrt{6}$.

In a triangle $ABC$,if $a = 2$,$B = 60^\circ$ and $C = 75^\circ$,then $b =$

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