As suggested by the first comment on the post here, I am trying to use law of sine to prove the result the greatest angle is opp. to greatest side. Let the triangle have angles $a,b,c$ opposite to side lengths $A,B,C$ then sine law:
$$ \frac{\sin a}{A} = \frac{\sin b}{B} = \frac{\sin c}{C} = \kappa$$
Rewriting this for side length:
$$ A = \frac{\sin a}{\kappa}$$
$$ B = \frac{\sin b}{\kappa}$$
$$ C = \frac{\sin c}{\kappa}$$
If an angle is in $\left[ 0 , \frac{\pi}{2} \right]$ then sine is strictly increasing in that domain and the greatest angle is opposite to greatest side. However, the above idea isn't applicable if the angle is in $\left[ \frac{\pi}{2} , \pi\right] $ because sine is no longer a strictly increasing function. Hence, I want to know how to extend the proof for cases when we have an obtuse triangle.
Best Answer
Your already showed it for acute angled and right angled triangles.
For obtuse angled triangle, if $A$ is the obtuse angle,
$a = k \sin A = k \sin (180^0-A) = k \sin(B+C)$
As $B+C \lt \frac{\pi}{2}$ and sin function is increasing in $[0, \frac{\pi}{2}]$,
we must have $ \ \sin(B+C) \gt \sin B$ and $\sin(B+C) \gt \sin C$.
Hence $a$ is the greatest side of the triangle.