Question

In a triangle ABC, the internal bisector of the $\angle ABC$ meets $AC$ at $D$. If $AB=4,AC=3$ and $\angle A={60}^{\circ },$ then the length of $AD$ is

• A. $2\sqrt{3}$
• B. $\frac{12}{4+\sqrt{13}}$
• C. $\frac{15\sqrt{3}}{8}$
• D. None of these

Answer 1

Answer: (B)

$\frac{12}{4+\sqrt{13}}$

Using Sine law in $\Delta ABD$;

$\frac{BD}{\sin 60°}=\frac{AD}{\sin B/2}⟹BD.\sin B/2=AD.\sin 60°$ (1)

Using Sine law in $\Delta BCD$;

$\frac{BD}{\sin C}=\frac{3-AD}{\sin B/2}⟹BD.\sin B/2=(3-AD).\sin C$ (2)

Using Sine law in $\Delta ABC;$

$\frac{a}{\sin 60°}=\frac{4}{\sin C}⟹\sin C=\frac{4}{a}\sin 60°$ (3)

BY equation 1 and 2 and 3; we get

$AD\sin 60°=(3-AD)\frac{4}{a}\sin 60°$

$⟹AD.a=(3-AD)4$ (4)

Using,

$\sin A/2=\sqrt{\frac{(s-b)(s-c)}{bc}}$

$⟹\sin 30°=\sqrt{\frac{(s-3)(s-4)}{12}}$

$⟹\frac{1}{2}=\sqrt{\frac{s^2-7s+12}{12}}$

Squaring on both sides;

$\frac{1}{4}=\frac{s^2-7s+12}{12}⟹s^2-7s+9=0$

$⟹s=\frac{7\pm \sqrt{1}3}{2}⟹2s=7+\sqrt{1}3$

$2s=a+b+c$

$⟹2s=a+7=7+\sqrt{1}3$

$⟹a=\sqrt{1}3$

Then; equation 4

$AD.\sqrt{1}3=12-4AD$

$⟹AD(4+\sqrt{1}3)=12$

$⟹AD=\frac{12}{4+\sqrt{1}3}$