Never thought about it

Informal approach: $0^\circ$ and $90^\circ$ match, so it's $90^\circ$ .

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Formal approach:

Let $\sin A$ be $s$ .

$s + \sqrt{1-s^2} = 1$

$\sqrt{1-s^2} = 1-s$

$1-s^2 = s^2 - 2s + 1$

$2s^2 - 2s = 0$

$2s(s-1)=0$

$s$ = 0 or $s$ = 1

$A$ = $0^\circ$ or $A$ = $90^\circ$

we have the identity sin^2A+cos^2A=1 hence 1=sinA+cosA 1=1/2+1/2 sin(1/2)=30 degrees cos(1/2)=60 degrees therefore,30+60=90 degrees !