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Math Help - finding sine, cosine, and tangent of theta

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    finding sine, cosine, and tangent of theta

    I have a right triangle with x, y, and z is the hypotenuse. It has angles A and B.The given information is that z=17, A=theta, and B=theta. I used the pythagorean theorem to deduce that the sides are equal length at sqrt(289/2). I am to find sin of theta which is the opposite side over the hypotenuse, and is sqrt(289/2)/17 which reduces to 1/sqrt(2) i am unsure how it goes from sqrt(289/2)/17 to 1/sqrt(2) could someone please explain?
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    Quote Originally Posted by sydewayzlocc View Post
    I have a right triangle with x, y, and z is the hypotenuse. It has angles A and B.The given information is that z=17, A=theta, and B=theta. I used the pythagorean theorem to deduce that the sides are equal length at sqrt(289/2). I am to find sin of theta which is the opposite side over the hypotenuse, and is sqrt(289/2)/17 which reduces to 1/sqrt(2) i am unsure how it goes from sqrt(289/2)/17 to 1/sqrt(2) could someone please explain?
    Hi sydewayzlocc,

    I'd be happy to explain.

    Two ways...

    First, you have the legs at \sqrt{\frac{289}{2}}

    That simplifies to \frac{\sqrt{289}}{\sqrt{2}}=\frac{17}{\sqrt{2}}

    \sin \theta = \dfrac{\frac{17}{\sqrt{2}}}{17}

    \sin \theta = \dfrac{17}{\sqrt{2}} \times \dfrac{1}{17}=\dfrac{1}{\sqrt{2}}

    You would probably want to "rationalize" that answer: \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}


    Or you could rationalize the lengths of the two legs first:

    \sqrt{\frac{289}{2}\times \frac{2}{2}} = \sqrt {\frac{17^2 \times 2}{4}}=\frac{17\sqrt{2}}{2}

    \sin \theta = \dfrac{\frac{17\sqrt{2}}{2}}{17}

    \sin \theta = \dfrac{\sqrt{2}}{2} which is the same as \dfrac{1}{\sqrt{2}}

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