What is the value of the cotangent $\frac{(B-C)}{2}$?

Question

In a triangle $ABC$, it is known that its area is $ 4\sqrt{3} \ \mathrm{m^2}, \angle A = 60^\circ $ and the internal bisector of the angle $A$ measures $ \frac{8\sqrt{3}}{5} \ \mathrm m$. If $B > C$, calculate $\cot\frac{(B-C)}{2}$.

I try use $$S =\frac{1}{2}bc\sin A\rightarrow 4\sqrt3 =\frac{1}{2}bc\frac{\sqrt3}{2}\\ \therefore \boxed{bc=16} \\ a^2=b^2+c^2-\frac{1}{2}bc\cos A = b^2+c^2-\frac{1}{2}\cdot16\cdot\frac{1}{2}=b^2+c^2-4 \\ \therefore \boxed{a^2=b^2+c^2-4}$$ I stop here. Can someone help? (The solution is $\frac{4\sqrt3}{5})$

Answer 1

Let the bisector of A intercept BC at D.

$S =\frac{1}{2}bc\sin A, A=60^\circ \rightarrow \frac{1}{4}\sqrt3 b c=4\sqrt 3.b.c \therefore \color{green}{bc=16} \\$

$S=S_{\triangle ACD}+S_{\triangle ABD}=\frac 14(b+c)\cdot AD=\frac 14 \cdot \frac{8\sqrt{3}}{5}\cdot (b+c)=4\sqrt 3 \therefore \color{green}{b+c=10} \\ $

Given $B >C, \color{green}{b= 8, c=2}$.

Apply law of cosine given A,b,c in $\triangle ABC \to a=2\sqrt {13}$

Given $a,b,c$ now, $\\\cos C=\frac{7}{2\sqrt{13}}\approx0.971, \color{green}{C\approx14^\circ},\\ cos B=-\frac{1}{\sqrt{13}}\approx -0.278, \color{green}{B\approx 106^\circ} \\ \boxed {\cot \frac{(B-C)}{2}\approx \cot 46^\circ\approx 0.9656}$

Somehow, your solution is not consistent with this result.

Answer 2

Hint : Try to use tha fact that , the length of the internal angle bisector from vertex A equals (2bc/b+c)*cos(A/2) , and that cot(B-C/2)= (b+c/b-c)tan(A/2).