In a triangle $A B C$, 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 $ an A=$

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In a triangle $A B C$, 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 $	an A=$

Aicharya · Accepted Answer

Since $A, B, C$ are in A.P. Let $B=60^{\circ}$ $\begin{aligned} & \Rightarrow C=120-A \quad\left[\because A+B+C=180^{\circ}ight] \ & 	ext { Now, } \cos A+\cos B+\cos C=\frac{1+\sqrt{2}+\sqrt{3}}{2 \sqrt{3}} \ & \Rightarrow \cos A+\cos C=\frac{1+\sqrt{2}+\sqrt{3}}{2 \sqrt{3}}-\frac{\sqrt{2}}{2} \ & =\cos A+\cos (120-A)=\frac{1+\sqrt{3}}{2 \sqrt{2}} \ & \Rightarrow 2 \cos \left(\frac{A+120-A}{2}ight) \cos \left(\frac{A-120+A}{2}ight)=\cos 15^{\circ} \ & \Rightarrow \cos (A-60)=\cos 15^{\circ} \Rightarrow A=75^{\circ} \ & 	an A=	an 75^{\circ}=	an \left(45^{\circ}+30^{\circ}ight)=2+\sqrt{3} . \end{aligned}$

In a triangle $A B C$, if $A, B, C$ are in arithmetic progression and $\cos A+\cos B+\cos…

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