sin x = 15 / √5 .... but you can multiply by √5/√5 to rationalize the denominator
= 15√5 / 5
= 3√5
Now comes the problem. That number is greater than 1, so it can't be the sine of any real-valued angle. (You can extend the trig functions to complex numbers and get an answer, but you won't need that for years to come--if ever.)
Answers & Comments
sin x = 1/(csc x)
In this case, that's
sin x = 15 / √5 .... but you can multiply by √5/√5 to rationalize the denominator
= 15√5 / 5
= 3√5
Now comes the problem. That number is greater than 1, so it can't be the sine of any real-valued angle. (You can extend the trig functions to complex numbers and get an answer, but you won't need that for years to come--if ever.)
1 / sin Ө = √5 / 15
sin Ө = 15 / √5
is not true because sin Ө ≤ 1
Csc = 1/Sin
Hence
Csc(Th) = 1/ Sin(Th) = sqrt(5)/5
Inverting
Sin(Th) = 15 / Sqrt(5)
Sin(Th) = 15 sqrt(5) / Sqty(5)(sqrt(5)
Sin(th) = 15sqrt(5) / 5
Sin(Th) = 3sqrt(5) > 1 Hence NOT resolved.