Question

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

• 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

Answer: (B)

$(3+\sqrt{5})\text{cosec}C$

$\frac{4}{\sin 18}=\frac{c}{\sin 126}$

$\frac{4}{\sin 18}=\frac{c}{\cos 36}$

$\Rightarrow c=4\left(\frac{1-2{\sin }^{2}18}{\sin 18}\right)$

$\frac{c}{\sin C}=2R$

$\Rightarrow R=\frac{1}{2}\frac{c}{\sin C}$

Now,

$\Rightarrow R=2\left(\frac{1-2{\sin }^{2}18}{\sin 18}\right)\text{cosec}C$

We know that

$\sin 18=\frac{1}{4}(\sqrt{5}-1)$

Therefore,

$\Rightarrow R=(\sqrt{5}+3)\text{cosec}C$

Hence, option B.