Question

In a $\Delta ABC,\angle A={120}^{\circ }$. The bisector of $A$ cut $BC$ at point $D$, such that length of $BD$ is twice of $CD$. If $AD=10$ unit and $BC=k\sqrt{7}$ unit, then value of $k$ is

Answer 1

$\text{Let}CD=a\Rightarrow BD=2a\text{and}\angle ADC=x\text{Applying sine law in}\Delta ACD,\frac{AC}{\sin x}=\frac{a}{\sin {60}^{\circ }}...\left(1\right)\text{Applying sine law in}\Delta ABD,\frac{AB}{\sin (180-x)}=\frac{2a}{\sin {60}^{\circ }}\Rightarrow \frac{AB}{\sin x}=\frac{2a}{\sin {60}^{\circ }}...\left(2\right)\text{From eqn(1) and (2), we get}AB=2AC\text{Applying cosine law in}\Delta ACD,a^2=b^2+{10}^2-2\times b\times 10\times \cos {60}^{\circ }\Rightarrow a^2=b^2+100-10b...\left(3\right)\text{Applying cosine law in}\Delta ABD,(2a{)}^2=(2b{)}^2+{10}^2-2\times 2b\times 10\times \cos {60}^{\circ }\Rightarrow 4a^2=4b^2+100-20b...\left(4\right)\text{From eqn(3) and (4), we get}a=5\sqrt{7},b=15BC=3a=15\sqrt{7}\Rightarrow k=15$