A special triangle and relation

let B+C=180-A obviously Then rewrite the equation like this ==> sinA=tan[(180-A)/2]

using the half angle formula for tangent, you'll end up with ==> sinA = -sinA/(-1+cosA)

with that, you'll notice that cosA must be equal to zero for the equation to be possible.

it will give you two answers A=90 or A=270; but of course, an angle of a triangle couldn't be more than or equal to 180 degrees if you haven't noticed that, you probably should study triangles again , so the answer is obviously A=90 degrees.

$sinA$ = 2 $sin\frac{A}{2}$ $cos\frac{A}{2}$ . Because the angles are in a triangle, their sum divided by 2 is 90 degrees, so $tg\frac{B+C}{2}$ = $\frac{cos\frac{A}{2}}{sin\frac{A}{2}}$ . With these two relations and the given relation, we can find that $sin\frac{A}{2}$ = I $\frac{\sqrt{2}}{2}$ I. We can conclude that A = 90 or 270. Because a triangle can't have angle equal to 270 degrees, then the solution is 90. =D