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Trig Help
Hi, I need help finding the sin of the angle alpha in standard postion whose terminal side contains the given point. (4,6).
Here's what I have so far.
r = sqrt{(4)^2 + (6)^2} = sqrt{52}
x = 4, y= 6, r= sqrt{52}
sin(alpha) = (6/sqrt(52))
= (6/sqrt(52)*sqrt(52)/sqrt(52)
=6(sqrt(13*2*2) / 52)
The answer ends up being sin(alpha)= {3(sqrt(13)/13}
And that is where I am getting lost... It's simpler algebra that I should already know, but I just don't remember. Here's an image that might be easier on the eyes...
The box is around the answer I should get to. Thanks for any help.

You're good.
Let us continue,
sin(alpha) = (6/sqrt(52))
Rationalize the denominator,
= (6/sqrt(52)*sqrt(52)/sqrt(52)
= 6(sqrt(13*2*2) / 52)
Get the 2*2 out of the sqrt sign,
= 6[2sqrt(13)] / 52
= 12[sqrt(13)] / 52
Reduce it to its lowest term, divide both numerator and denominator by 4,
= 3[sqrt(13)] / 13 answer.

:) Thank you very much! Your help is much appreciated