# How to prove that this triangle is right-angled?

## Letters a, b and c are for the angles of a triangle. How to prove that if $\cot \left(\frac{a}{2}\right) = \frac{\sin b + \sin c}{\sin a}$, than is the triangle right-angled

Dec 10, 2017

If you have all three side lengths, to be right angled the triangle must obey Pythagorus's theorem.

#### Explanation:

eg.
if triangle has side lengths of 3, 4 and 5;
${3}^{2} + {4}^{2} = {5}^{2}$
$9 + 16 = 25$
Hence triangle is right angled.
If however, the triangle has side lengths of 3, 4 and 6;
${3}^{2} + {4}^{2} \ne {6}^{2}$
$9 + 16 \ne 36$
and triangle is NOT right angled.

Hope it helps :)

Dec 10, 2017

Kindly refer to a Proof given in the Explanation.

#### Explanation:

Given $\Delta$ has $3$ angles $a , b , c .$

$\therefore a + b + c = \pi , \mathmr{and} , b + c = \pi - a .$

Also, we know that,

$\sin b + \sin c = 2 \sin \left(\frac{b + c}{2}\right) \cos \left(\frac{b - c}{2}\right) .$

Now, $\cot \left(\frac{a}{2}\right) = \frac{\sin b + \sin c}{\sin} a ,$

$\Rightarrow \cot \left(\frac{a}{2}\right) = \frac{2 \sin \left(\frac{b + c}{2}\right) \cos \left(\frac{b - c}{2}\right)}{\sin} a ,$

$\Rightarrow \cot \left(\frac{a}{2}\right) = \frac{2 \sin \left(\frac{\pi - a}{2}\right) \cos \left(\frac{b - c}{2}\right)}{\sin} a ,$

$= \frac{2 \sin \left(\frac{\pi}{2} - \frac{a}{2}\right) \cos \left(\frac{b - c}{2}\right)}{\sin} a ,$

$= \frac{2 \cos \left(\frac{a}{2}\right) \cos \left(\frac{b - c}{2}\right)}{2 \sin \left(\frac{a}{2}\right) \cos \left(\frac{a}{2}\right)} .$

$\therefore \cot \left(\frac{a}{2}\right) = \cot \left(\frac{a}{2}\right) \left\{\cos \frac{\frac{b - c}{2}}{\cos} \left(\frac{a}{2}\right)\right\} , i . e . ,$

$\cot \left(\frac{a}{2}\right) - \cot \left(\frac{a}{2}\right) \left\{\cos \frac{\frac{b - c}{2}}{\cos} \left(\frac{a}{2}\right)\right\} = 0.$

$\cot \left(\frac{a}{2}\right) \left[1 - \left\{\cos \frac{\frac{b - c}{2}}{\cos} \left(\frac{a}{2}\right)\right\}\right] = 0.$

$\therefore \cot \left(\frac{a}{2}\right) = 0 , \mathmr{and} , 1 - \left\{\cos \frac{\frac{b - c}{2}}{\cos} \left(\frac{a}{2}\right)\right\} = 0.$

$\therefore \cot \left(\frac{a}{2}\right) = 0 , \mathmr{and} , 1 = \left\{\cos \frac{\frac{b - c}{2}}{\cos} \left(\frac{a}{2}\right)\right\} , i . e . ,$

$\therefore \cot \left(\frac{a}{2}\right) = 0 , \mathmr{and} , \cos \left(\frac{b - c}{2}\right) = \cos \left(\frac{a}{2}\right) ,$

$\therefore \cos \frac{\frac{a}{2}}{\sin} \left(\frac{a}{2}\right) = 0 , \mathmr{and} , \cos \left(\frac{b - c}{2}\right) = \cos \left(\frac{a}{2}\right) ,$

$\therefore \cos \left(\frac{a}{2}\right) = 0 = \cos \left(\frac{\pi}{2}\right) , \mathmr{and} , \cos \left(\frac{b - c}{2}\right) = \cos \left(\frac{a}{2}\right) .$

Taking into a/c that $a , b , c$ are angles of a $\Delta ,$ we have,

$\therefore \frac{a}{2} = \frac{\pi}{2} , \mathmr{and} , \frac{b - c}{2} = \frac{a}{2.}$

$\therefore a = \pi ,$ which is not possible, or, $b - c = a .$

 b-c=a, &, b+c=pi-a rArr 2b=pi, or, b=pi/2.

This proves that $\Delta$ is right-angled at $\angle b .$