The angles A, B and C of a triangle ABC are i | vedclass

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(A) Given that angles $A, B, C$ are in $A.P.$,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.Usin

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$2\sqrt{3}$

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(A) Given that angles $A, B, C$ are in $A.P.$,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.Given $\frac{a}{b} = \frac{1}{\sqrt{3}}$,we have $\frac{\sin A}{\sin B} = \frac{1}{\sqrt{3}}$.Since $B = 60^{\circ}$,$\sin B = \frac{\sqrt{3}}{2}$,so $\sin A = \frac{1}{\sqrt{3}} 	imes \frac{\sqrt{3}}{2} = \frac{1}{2}$,which means $A = 30^{\circ}$.Then $C = 180^{\circ} - (30^{\circ} + 60^{\circ}) = 90^{\circ}$.Given $c = 4$,we have $a = c \sin A = 4 \sin 30^{\circ} = 2$ and $b = c \sin B = 4 \sin 60^{\circ} = 2\sqrt{3}$.The area of the triangle is $\Delta = \frac{1}{2} ab \sin C = \frac{1}{2} 	imes 2 	imes 2\sqrt{3} 	imes \sin 90^{\circ} = 2\sqrt{3} 	ext{ sq. cm}$.

The angles $A, B$ and $C$ of a triangle $ABC$ are in $A.P.$ and $a : b = 1 : \sqrt{3}$. If $c = 4 \text{ cm}$,then the area (in $\text{sq. cm}$) of this triangle is:

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