Question

In a $\Delta ABC,$ the bisector of $\angle A$ meets the side $BC$ at point $D$, such that $BD=2CD$. If $AD=10$ unit and $BC=15\sqrt{7}$ unit, then the value of $\angle A$ is

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${120}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is C ${120}^{\circ }$


$\text{Let}CD=a\Rightarrow BD=2aBC=BD+CD\Rightarrow 15\sqrt{7}=3a\Rightarrow a=5\sqrt{7}\text{Let}\angle ADC=x,\angle BAC=2y$

$\text{Applying sine law in}\Delta ACD,\frac{AC}{\sin x}=\frac{a}{\sin y}...\left(1\right)\text{Applying sine law in}\Delta ABD,\frac{AB}{\sin (180-x)}=\frac{2a}{\sin y}\Rightarrow \frac{AB}{\sin x}=\frac{2a}{\sin y}...\left(2\right)\text{From eqn(1) and (2), we get}AB=2AC=2b\left(\text{say}\right)$

$\text{Applying cosine law in}\Delta ACD,a^2=b^2+{10}^2-2\times b\times 10\times \cos y\Rightarrow (5\sqrt{7}{)}^2=b^2+100-20b\cos y\Rightarrow 75=b^2-20b\cos y...(3)\text{Applying cosine law in}\Delta ABD,(2a{)}^2=(2b{)}^2+{10}^2-2\times 2b\times 10\times \cos y\Rightarrow 600=4b^2-40b\cos y...(4)$

$\text{Now, eqn(4) - eqn(3)}\times 2,\text{gives}b=15\text{From eqn(3),}75={15}^2-20\times \cos y\Rightarrow y={60}^{\circ }\therefore \angle A={120}^{\circ }$