<s>[INST] If $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$, then is it unique and its area? Given that $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$. Then would the $\triangle ABC$ be unique? If so how easily we may find its area or otherwise the maximum possible area? 
Any help is appreciated :)
 [/INST] Note $A+B+C=\pi,\cos {(A+B+C)}=-1 \implies 1-\cos(A)\sin(B)\sin(C)-\sin(A)\cos(B)\sin(C)-\sin(A)\sin(B)\cos(C)+\cos(A)\cos(B)\cos(C)=0$
let $D=2R \implies D \ge 3 $, we have $\sin A= \sqrt{1-\cos^2 A}=\dfrac{\sqrt{D^2-1^2}}{D},\sin B=\dfrac{\sqrt{D^2-2^2}}{D},\sin C=\dfrac{\sqrt{D^2-3^2}}{D}$
$D^3+1*2*3=3\sqrt{D^2-1^2}\sqrt{D^2-2^2}+2\sqrt{D^2-1^2}\sqrt{D^2-3^2}+1*\sqrt{D^2-3^2}\sqrt{D^2-2^2} \iff (2D+12)*\sqrt{(D^2-9)}\sqrt{(D^2-4)}=D^4-12D^2+12D+72 \iff D^3-14D-12=0 \iff D=4.1131$
which means the triangle is unique. 
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