In a triangle ABC,if c^2-a^2=b(sqrt{3}c-b) an | vedclass

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(D) Given equations are: $c^2 - a^2 = \sqrt{3}bc - b^2$  $(1)$ $b^2 - a^2 = c^2 - ac$  $(2)$ From $(1)$,$a^2 = c^2 + b^2 - \sqrt{3}bc$. By the Law of

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(D) Given equations are: $c^2 - a^2 = \sqrt{3}bc - b^2$  $(1)$ $b^2 - a^2 = c^2 - ac$  $(2)$ From $(1)$,$a^2 = c^2 + b^2 - \sqrt{3}bc$. By the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$. Comparing these,$2bc \cos A = \sqrt{3}bc \implies \cos A = \frac{\sqrt{3}}{2} \implies A = 30^{\circ}$. From $(2)$,$a^2 = b^2 - c^2 + ac$. By the Law of Sines,$a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$. Substituting these into $(2)$: $\sin^2 B - \sin^2 A = \sin^2 C - \sin A \sin C$. Using $A = 30^{\circ}$,$\sin A = 1/2$. $\sin^2 B - 1/4 = \sin^2 C - \frac{1}{2} \sin C$. Since $B = 180^{\circ} - (A+C) = 150^{\circ} - C$,$\sin B = \sin(150^{\circ}-C) = \sin 150^{\circ} \cos C - \cos 150^{\circ} \sin C = \frac{1}{2} \cos C + \frac{\sqrt{3}}{2} \sin C$. Substituting this and solving for $C$ leads to $C = 90^{\circ}$.

In a triangle $ABC$,if $c^2-a^2=b(\sqrt{3}c-b)$ and $b^2-a^2=c(c-a)$,then $\angle ACB=$ (in $^{\circ}$)

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