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

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(C) Since $A, B, C$ are in $AP$,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we have $A + C = 180^{\circ} - B$. Substituting this into the $A

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(C) Since $A, B, C$ are in $AP$,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we have $A + C = 180^{\circ} - B$. Substituting this into the $AP$ condition: $2B = 180^{\circ} - B$ $\Rightarrow 3B = 180^{\circ}$ $\Rightarrow B = 60^{\circ}$. Using the sine rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$. Given $\frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}$,we have $\frac{\sin B}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$. Substituting $B = 60^{\circ}$: $\frac{\sin 60^{\circ}}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$ $\Rightarrow \frac{\sqrt{3}/2}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$. $\sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow C = 45^{\circ}$. Finally,$A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 75^{\circ}$.

The angles $A, B$ and $C$ of a triangle $ABC$ are in $AP$. If $b: c = \sqrt{3}: \sqrt{2}$,then the angle $A$ is (in $^{\circ}$)

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