In $∆$ABC, AD and BE are the medians drawn through the angular points A and B respectively. $\mathrm{\angle }\mathrm{DAB}=2\mathrm{\angle }\mathrm{ABE}={36}^{\circ }\text{ and }\mathrm{AD}=6\text{ units }$ then circum radius of the triangle equal to

By Applying sine rile in $∆$AGB

$\begin{array}{l}\frac{4}{\sin 18°}=\frac{c}{\sin 126°}(\because G divides \overline{AD} in the ratio 2:1 then AG=4,GD=2)\\ \Rightarrow \frac{4}{\sin 18°}=\frac{c}{\cos 36°}\\ \Rightarrow \mathrm{c}=4\left(\frac{\cos 36°}{\sin 18°}\right)=4\left(\frac{{\displaystyle \frac{\sqrt{5}+1}{4}}}{\frac{\sqrt{5}-1}{4}}\right)(\because \sin 18°=\frac{1}{4}(\sqrt{5}-1),\cos 36°=\frac{1}{4}(\sqrt{5}+1\left)\right)\\ \Rightarrow c=4\left(\frac{\sqrt{5}+1}{\sqrt{5}-1}\right)=4\left(\frac{{(\sqrt{5}+1)}^2}{(\sqrt{5}-1)(\sqrt{5 }+1)}\right)=6+2\sqrt{5}\\ Now from \sin e rule\frac{c}{\sin \mathrm{C}}=2R\\ \Rightarrow \mathrm{R}=\frac{\mathrm{c}}{2\sin \mathrm{C}}\\ \Rightarrow \mathrm{R}=\left(\frac{6+2\sqrt{5}}{2}\right)\mathrm{cosec}\mathrm{C}\\ \\ \Rightarrow \mathrm{R}=(\sqrt{5}+3)\mathrm{cosec}\mathrm{C}\end{array}$