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 ? 

The cosine formula states
$$a^2 = b^2+ c^2 – 2bc \cos A$$
and since
$$\cos A = \frac{ \cot^2 (A/2) – 1}{ \cot^2(A/2)+1}$$
we have
$$a^2 = b^2 + c^2 -2bc \frac{ (b+c)^2 – a^2}{(b+c)^2 + a^2},$$
which reduces to $a^4 = (b^2 – c^2)^2.$
And so taking the square root $a^2 =b^2 – c ^2,$ where $b$ is the larger side.
Hence $b^2 = c^2+a^2$ and so we have a right-angled triangle.
And so it's the angle opposite the larger of $b$ and $c$ which is the right-angle.
