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Life Group Title

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

  • 2 years ago
  • 2 years ago

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  1. Life Group Title
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    |dw:1352794320204:dw|

    • 2 years ago
  2. maym0Re97 Group Title
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    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)

    • 2 years ago
  3. maym0Re97 Group Title
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    |dw:1352794508956:dw|

    • 2 years ago
  4. Life Group Title
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    i got 5/4 but thats not one of the choices. do you see where i went wrong?

    • 2 years ago
  5. Life Group Title
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    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

    • 2 years ago
  6. maym0Re97 Group Title
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    Hrm. Ok. Uhhh what are all the choices?

    • 2 years ago
  7. Life Group Title
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    sqrt 3/5, sqrt 13/4, 3/5, 4/5

    • 2 years ago
  8. maym0Re97 Group Title
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    it's sqrt 13/4

    • 2 years ago
  9. hartnn Group Title
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    alternative way: sin^2 x + cos^2 x = 1

    • 2 years ago
  10. maym0Re97 Group Title
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    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.

    • 2 years ago
  11. maym0Re97 Group Title
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    Woops for the first equation it should be sqrt3, not sqrt 13

    • 2 years ago
  12. maym0Re97 Group Title
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    lol and it should be 13=x^2, not 13=x. I hate the equation thing.

    • 2 years ago
  13. Life Group Title
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    lol so what i did was correct right?

    • 2 years ago
  14. maym0Re97 Group Title
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    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.

    • 2 years ago
  15. maym0Re97 Group Title
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    So the final answer you got SHOULD be \[\sin=\sqrt{13}/4\]

    • 2 years ago
  16. Life Group Title
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    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?

    • 2 years ago
  17. maym0Re97 Group Title
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    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.

    • 2 years ago
  18. Life Group Title
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    ughh didnt see that, thanks lol !

    • 2 years ago
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