Problem related to Incircle of a triangle

Question

Given a $\triangle{ABC}$, Draw an incircle touching sides $AB,BC,AC$ at points $D,E,F$ respectively.Given $\angle{A}=60°$, and lengths of $AD=5$ cm and $DB=3$cm.

Calculate the side length of $BC$.

My Attempt:

I know that AD=AF, CF=CE and BD=BE (Tangents to a circle from a common point)

I can also write,

$$AD=r\cot{\frac{A}{2}}$$ $$BE=r\cot{\frac{B}{2}}$$ $$CF=r\cot{\frac{C}{2}}$$

Putting AD=$5$ cm, and $\angle{A}=60°$.

I calculated $$r=\frac{5}{\sqrt{3}}$$

Then calculated $$\cot{\frac{B}{2}}={\frac{3\sqrt{3}}{5}}$$

How do I proceed next?

I have in mind to use something like, $$\cot{\frac{A}{2}}+\cot{\frac{B}{2}}+\cot{\frac{C}{2}}=\cot{\frac{A}{2}}\cot{\frac{B}{2}}\cot{\frac{C}{2}}$$

Am I solving this right?

Answer 1

Hint: Use that

$$a^2=8^2+(2+a)^2-2\cdot 8\cdot (a+2)\cos(60^{\circ})$$ solve this equation for $a$

Answer 2

Suppose the incenter is $P$. Then $r=DP$. $$\require{cancel}r=\tan 30^\circ\cdot5=\frac{5\sqrt3}{3}$$ Then: $$\angle ABP=\arctan\left(\frac{5\sqrt3}{9}\right)\implies\angle B=2\angle ABP=2\arctan\left(\frac{5\sqrt3}{9}\right)\\ \angle C=120^\circ-2\arctan\left(\frac{5\sqrt3}{9}\right)\approx 32.20^\circ\\ \frac{\sin A}{BC}=\frac{\sin C}{AB}\implies BC=\frac{(\sin A)\cdot AB}{\sin C}=\frac{4\sqrt3}{\sin\left(120^\circ-2\arctan\left(\frac{5\sqrt3}{9}\right)\right)}=\frac{\cancel{4\sqrt3}}{\frac{\cancel{4 \sqrt{3}}}{13}}=13$$