If we have a right triangle, which we know the Hypotenuse only, and let's assume H=3, A (adjacent) and O (opposite) are both unknown, as well as both $\theta_{1}$ and $\theta_{2}$ angles are unknown. But we know the angle $\theta_{3} = 90°$ as always, so we get the following two equations to solve for Opposite side and Adjacent side:
$\sin{(\frac{O}{3})} = 90°$
$\cos{(\frac{A}{3})} = 90°$,
which; we then need to solve for A and O...
By using arcus sinus, we get the following:
$\arcsin{(\sin{(\frac{O}{3})})} = \arcsin{(90°)} \implies O = 3 \cdot \arcsin{(90°)}$,
But arc sine can-not work with degrees, only with radians where $-1 > x < 1$, so if we try to convert $90°$ into radians, we must multiply by $π$ and then divide by $180$, which we get the following:
$\frac{90π}{180} \approx 1.57$,
which the following dominates:
$-1 > x > 1$,
while we need $-1 > x < 1$...
So... is there any solution for that???
Thanks for the answers.
Your question as stated cannot be answered, because there are many right triangles with hypotenuse 3, with different values of the lengths of the sides. Also, you have your input and output reversed in that sin(90º) = 1, sin(30º) = 1/2 and so forth. Trigonometric functions are defined for degrees and radians. It is when you do calculus that the difference becomes important.