The angles of triangle ABC are in A.P. and b | vedclass

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(D) Let the angles be $A-d, A, A+d$. Since the sum of angles is $180^\circ$, $(A-d) + A + (A+d) = 180^\circ \implies 3A = 180^\circ \implies A = 60^\c

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(D) Let the angles be $A-d, A, A+d$. Since the sum of angles is $180^\circ$, $(A-d) + A + (A+d) = 180^\circ \implies 3A = 180^\circ \implies A = 60^\circ$.Using the Sine Rule, $\frac{b}{\sin B} = \frac{c}{\sin C} \implies \frac{\sin B}{\sin C} = \frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}$.Since $A = 60^\circ$, $B+C = 120^\circ$, so $C = 120^\circ - B$.$\frac{\sin B}{\sin(120^\circ - B)} = \frac{\sqrt{3}}{\sqrt{2}} \implies \sqrt{2} \sin B = \sqrt{3} (\sin 120^\circ \cos B - \cos 120^\circ \sin B)$.$\sqrt{2} \sin B = \sqrt{3} (\frac{\sqrt{3}}{2} \cos B + \frac{1}{2} \sin B) = \frac{3}{2} \cos B + \frac{\sqrt{3}}{2} \sin B$.$(\sqrt{2} - \frac{\sqrt{3}}{2}) \sin B = \frac{3}{2} \cos B \implies 	an B = \frac{3}{2\sqrt{2} - \sqrt{3}}$.Alternatively, using $b^2 = a^2 + c^2 - 2ac \cos B$, we find $B = 75^\circ$ and $C = 45^\circ$. Thus, $\angle A = 60^\circ$ is incorrect based on the given ratio; re-evaluating: if $A, B, C$ are in $A$.$P$., $2B = A+C$. Since $A+B+C = 180^\circ$, $3B = 180^\circ \implies B = 60^\circ$. Then $A+C = 120^\circ$. Using $\frac{\sin A}{\sin C} = \frac{a}{c}$, this does not match. Given the options, the question implies $A, B, C$ are in $A$.$P$. such that $B=60^\circ$. The correct calculation for $\angle A$ with $b:c = \sqrt{3}:\sqrt{2}$ and $B=60^\circ$ leads to $\angle A = 75^\circ$.

The angles of $\triangle ABC$ are in $A$.$P$. and $b : c = \sqrt{3} : \sqrt{2}$, then $\angle A =$ (in $^\circ$)

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