What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ? 

OK, first rewrite it as
$$ \cos(A/2) = \frac{b+c}{a} \sin(A/2) $$
Multiply both sides by $2 \cos(A/2)$ 
$$ 2 \cos^2(A/2) = \frac{b+c}{a} \sin(A) $$
Now use the rule $a/\sin(A)=b/\sin(B)=c/\sin(C)$ to transform this to:
$$ 2 \cos^2(A/2) = \sin(B)+\sin(C) $$
Using Simpson's rule on the second hand of the equation to sum the sines gives:
$$ 2 \cos^2(A/2) = 2\sin((B+C)/2)\cos((B-C)/2) $$
Since in a triangle $A+B+C=\pi$ we have $ 2\sin((B+C)/2) = 2\sin((\pi-A)/2) = 2\cos(A/2) $ and thus
$$ \cos(A/2) = \cos((B-C)/2) $$
Or $A+C=B$ (If $B>C$). Combining this with $A+B+C=\pi$ this gives $B=\pi/2$.
