In a triangle ABC,if A, B, C are in arithmeti | vedclass

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

Vedclass · Accepted Answer

$2 + \sqrt{3}$

Vedclass · Answer

(B) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.Substituting $B = 60^{\circ}$ into the given equation: $\cos A + \cos 60^{\circ} + \cos C = \frac{1 + \sqrt{2} + \sqrt{3}}{2 \sqrt{2}}$.$\cos A + \cos C = \frac{1 + \sqrt{2} + \sqrt{3}}{2 \sqrt{2}} - \frac{1}{2} = \frac{1 + \sqrt{2} + \sqrt{3} - \sqrt{2}}{2 \sqrt{2}} = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$.Using the sum-to-product formula: $2 \cos \left( \frac{A+C}{2} ight) \cos \left( \frac{A-C}{2} ight) = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$.Since $A+C = 120^{\circ}$,$\frac{A+C}{2} = 60^{\circ}$,so $2 \cos 60^{\circ} \cos \left( \frac{A-C}{2} ight) = \cos \left( \frac{A-C}{2} ight) = \frac{1 + \sqrt{3}}{2 \sqrt{2}} = \frac{1}{2 \sqrt{2}} + \frac{\sqrt{3}}{2 \sqrt{2}} = \cos 75^{\circ}$.Thus,$\frac{A-C}{2} = 15^{\circ}$,which implies $A-C = 30^{\circ}$.Solving $A+C = 120^{\circ}$ and $A-C = 30^{\circ}$ gives $2A = 150^{\circ}$,so $A = 75^{\circ}$.Finally,$	an A = 	an 75^{\circ} = 	an(45^{\circ} + 30^{\circ}) = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$.

In a triangle $ABC$,if $A, B, C$ are in arithmetic progression and $\cos A + \cos B + \cos C = \frac{1 + \sqrt{2} + \sqrt{3}}{2 \sqrt{2}}$,then $\tan A =$

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