Question

The ratio of the lengths of the sides $BC$ and $AC$ of $\Delta ABC$ to the radius of circumscribed circle are equal to $2$ and $1.5$ respectively. The ratio of length of the bisectors of the interior angles $B$ and $C$ is

• A. $\frac{7(\sqrt{7}-1)}{9\sqrt{2}}$
• B. $\frac{7(\sqrt{7}+1)}{9\sqrt{2}}$
• C. $\frac{7(\sqrt{7}-1)}{9\sqrt{2}}$
• D. $\text{None of these}$

Answer 1

The correct option is A $\frac{7(\sqrt{7}-1)}{9\sqrt{2}}$


$\frac{a}{R}=2,\frac{b}{R}=1.5=\frac{3}{2}\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\Rightarrow \sin B=\frac{3}{4},\sin A=1\Rightarrow A={90}^{\circ }$
$\therefore c=\sqrt{a^2-b^2}=\frac{\sqrt{7}}{2}R$
$BD=\sqrt{A{B}^2+A{D}^2}$
$AD=AC.\frac{AB}{AB+BC}=\frac{bc}{a+c}$
$AE=\frac{bc}{a+b}$
$CE=\sqrt{A{C}^2+A{E}^2}$

$\text{Putting the values of a, b and c, we get}$
$AD=\frac{3\sqrt{7}R}{2(4+\sqrt{7})},AE=\frac{3\sqrt{7}R}{14}$
$BD=\frac{2\sqrt{7}R}{\sqrt{7}+1},CE=\frac{3\sqrt{2}R}{\sqrt{7}}\Rightarrow \frac{BD}{CE}=\frac{7(\sqrt{7}-1)}{9\sqrt{2}}$