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Math Help - Had a test today.. Need to know if this is right.

  1. #1
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    Had a test today.. Need to know if this is right.

    There's a right triangle with hypotenuse 5. I forgot what term it used, but they said something about it being in between 0 and \frac{\pi}{2}

    x is adjacent to theta. Find sin2\theta I already know that sin2\theta=2sin\theta cos\theta, it's just the length of the sides I was having trouble with.

    Am I correct in assuming this is just your standard 3, 4, 5 triangle and the \theta = \frac{24}{25}

    There was another way I tried to solve it.. where y(opposite to theta) was = \sqrt{25-x^2} but I didn't think that was correct so I went with the assumption I mentioned above..
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  2. #2
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    Quote Originally Posted by veryverymuch View Post
    There's a right triangle with hypotenuse 5. I forgot what term it used, but they said something about it being in between 0 and \frac{\pi}{2}

    x is adjacent to theta. Find sin2\theta I already know that sin2\theta=2sin\theta cos\theta, it's just the length of the sides I was having trouble with.

    Am I correct in assuming this is just your standard 3, 4, 5 triangle and the \theta = \frac{24}{25}

    There was another way I tried to solve it.. where y(opposite to theta) was = \sqrt{25-x^2} but I didn't think that was correct so I went with the assumption I mentioned above..
    The sine and cosine of an angle in a 3-4-5 triangle is pretty simple to get, I would have thought. As is the substitution of the values into the double angle formula. If you say what answer you gave, someone can confirm or reject it.
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    Quote Originally Posted by mr fantastic View Post
    The sine and cosine of an angle in a 3-4-5 triangle is pretty simple to get, I would have thought. As is the substitution of the values into the double angle formula. If you say what answer you gave, someone can confirm or reject it.
    Sorry if I was unclear. The details the question gave:
    x is adjacent to theta
    5 is the hypotenuse
    It is a right triangle.

    The answer I got was \theta=\frac{24}{25} when solving sin2\theta

    What I want to know is if it WAS a 3 4 5 triangle even though the question didn't state it. The wording is the only thing that confused me.
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    Quote Originally Posted by veryverymuch View Post
    Sorry if I was unclear. The details the question gave:
    x is adjacent to theta
    5 is the hypotenuse
    It is a right triangle.

    The answer I got was \theta=\frac{24}{25} when solving sin2\theta

    What I want to know is if it WAS a 3 4 5 triangle even though the question didn't state it. The wording is the only thing that confused me.
    I think you will find that there are not too many right-triangles with a hypotenuse of length 5 other than a 3-4-5 triangle. I suggest you research Pythagorean triples ....

    By the way, \theta=\frac{24}{25} is not a correct statement. Surely you mean \sin(2\theta)=\frac{24}{25}.
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    Quote Originally Posted by mr fantastic View Post
    I think you will find that there are not too many right-triangles with a hypotenuse of length 5 other than a 3-4-5 triangle. I suggest you research Pythagorean triples ....

    By the way, \theta=\frac{24}{25} is not a correct statement. Surely you mean \sin(2\theta)=\frac{24}{25}.
    Yeah, I meant sin2\theta
    Thanks either way, though.
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