If A = 30^circ, c = 7sqrt{3} and C = 90^circ | vedclass

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(B) Given that $C = 90^\circ$,the triangle is a right-angled triangle.Using the sine rule,we have $\frac{a}{\sin A} = \frac{c}{\sin C}$.Substituting t

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$\frac{7\sqrt{3}}{2}$

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(B) Given that $C = 90^\circ$,the triangle is a right-angled triangle.Using the sine rule,we have $\frac{a}{\sin A} = \frac{c}{\sin C}$.Substituting the given values: $a = \frac{c \sin A}{\sin C}$.$a = \frac{7\sqrt{3} \sin 30^\circ}{\sin 90^\circ}$.Since $\sin 30^\circ = \frac{1}{2}$ and $\sin 90^\circ = 1$,we get:$a = \frac{7\sqrt{3} 	imes (1/2)}{1} = \frac{7\sqrt{3}}{2}$.

If $A = 30^\circ, c = 7\sqrt{3}$ and $C = 90^\circ$ in $\Delta ABC$,then $a = $

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