if x is a positive acute angle and cos x = radical 3/4, what is the exact value of sin x?

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if x is a positive acute angle and cos x = radical 3/4, what is the exact value of sin x?

Mathematics
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|dw:1352794320204:dw|
Mk so cos is adjacent over hypotenuse, |dw:1352794400088:dw|, now you just have to use Pythagorean theorem to solve for the last side. Sin is opposite over hypotenuse, so you're going to take the value you found from Pythagorean theorem and put that over 4 (the hypotenuse value)
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i got 5/4 but thats not one of the choices. do you see where i went wrong?
one of the choices are 4/5 but im not sure, because i found the leg using the pythagorean theorem and got b(squared) = 25 which makes b= 5 and then over the hypotenuse which gave me 5/4
Hrm. Ok. Uhhh what are all the choices?
sqrt 3/5, sqrt 13/4, 3/5, 4/5
it's sqrt 13/4
alternative way: sin^2 x + cos^2 x = 1
You shouldn't be getting 5 for the last leg, because Pyth, should give you \[4^{2}=\sqrt{13}^{2}+x ^{2}\], which then becomes \[16=3+x ^{2}\], then \[13=x\], so x = sqrt 13.
Woops for the first equation it should be sqrt3, not sqrt 13
lol and it should be 13=x^2, not 13=x. I hate the equation thing.
lol so what i did was correct right?
Uhh you did the right thing, you just did it wrong. You go the wrong answer from Pyth. Theorem by yeah you're supposed to use that and then put it over 4.
So the final answer you got SHOULD be \[\sin=\sqrt{13}/4\]
i have a (squared) and c(squared) which means im looking for b (squared) i solved and got 9 +b(squared)=16, subtracted on both sides by 9 and got b2 = 7, not 5 my mistake, where did u get 13 from?
Aiiight. The thing is, your side length isn't 3, it's the sqrt of 3. So sqrt of 3 squared is 3. So then you take 16, which is 4 sqared, and subtract 3 from that, giving you 13. Then you end up with 13 = x squared. Take the square root of both sides, and you get square root of 13. Put that over 4 (hypotenuse) for your final answer.
ughh didnt see that, thanks lol !

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