Question

In triangle $ABC,AD$ and $BE$ are the medians drawn through the angular points $A$ and $B$ respectively. If $\angle DAB=2\angle ABE={36}^0$ and $AD=6$ units, then circumradius of the triangle is equal to
(where $A,B,C$ are angles opposite to the sides $BC,CA,AB$ respectively)

• A. $(3-\sqrt{5})\text{cosec}C$
• B. $(3+\sqrt{5})\text{cosec}C$
• C. $2(3-\sqrt{5})\text{cosec}C$
• D. $2(3+\sqrt{5})\text{cosec}C$

Answer 1

The correct option is B $(3+\sqrt{5})\text{cosec}C$

Let $G$ be the centroid, then $AG=4$ and $\angle AGB={126}^{\circ }$
$\therefore$In $\Delta AGB,\frac{4}{\sin {18}^{\circ }}=\frac{c}{\sin {126}^{\circ }}$$\frac{4}{\sin {18}^{\circ }}=\frac{c}{\cos {36}^{\circ }}$

$\Rightarrow c=4\left(\frac{\cos {36}^{\circ }}{\sin {18}^{\circ }}\right)=6+2\sqrt{5}\cdots \left(i\right)$
$\because \sin {18}^{\circ }=\frac{1}{4}(\sqrt{5}-1),\cos {36}^{\circ }=\frac{1}{4}(\sqrt{5}+1)$
Now $\frac{c}{\sin C}=2R$
$\Rightarrow R=\frac{1}{2}\frac{c}{\sin C}$
Putting value of $c$ from equation $\left(i\right)$
$\Rightarrow R=(\sqrt{5}+3)\text{cosec}C$