Question

In a $\Delta ABC,\angle A={120}^{\circ }$. If the angle bisector of $A$ cut $BC$ at point $D$, such that length of $BD$ is twice of $CD$ and $AD=10$ unit, then $BC$ is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

• A. $5\sqrt{7}\text{units}$
• B. $15\sqrt{10}\text{units}$
• C. $17\sqrt{5}\text{units}$
• D. $15\sqrt{7}\text{units}$

Answer 1

The correct option is D $15\sqrt{7}\text{units}$


Let $CD=a\Rightarrow BD=2a\text{and}\angle ADC=x$
Using sine rule, we get
$AC=\frac{a}{\sin {60}^{\circ }}\sin x=\frac{2a\sin x}{\sqrt{3}}AB=\frac{2a}{\sin {60}^{\circ }}\sin (180-x{)}^{\circ }=\frac{4a\sin x}{\sqrt{3}}$
Let $AC=b\Rightarrow AB=2b$
Using cosine rule, we get
$a^2=b^2+100-20b\cos {60}^{\circ }\Rightarrow a^2=b^2+100-10b\cdots \left(1\right)4a^2=4b^2+100-40b\cos {60}^{\circ }\Rightarrow a^2=b^2+25-5b$
Using equation $\left(1\right)$, we get
$\Rightarrow b^2+100-10b=b^2+25-5b\Rightarrow b=15\Rightarrow a=\sqrt{225+25-75}=5\sqrt{7}\therefore BC=15\sqrt{7}\text{units}$