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Math Help - Question with special angles and Trig Identities

  1. #1
    Senior Member vaironxxrd's Avatar
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    Question with special angles and Trig Identities

    Hello again Everyone,

    I had a question I was able to finally solve by myself but I was wondering if the following method is correct.

    The problem is, cot\theta = \frac{\sqrt{3}}{3}

    How I solved it...
    tan \theta = \frac{1}{cot \theta}

    tan \theta = \frac{1}{\frac{\sqrt3}{3}} Which equals 1.732

    Since I knew the question asked about special triangles I did the following
    tan(30) = 0.57735
    tan(45) = 1
    tan(60) = 1.73205
    Therefore \theta = 60

    Is this an acceptable method or the "right" method?
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  2. #2
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    Re: Question with special angles and Trig Identities

    Quote Originally Posted by vaironxxrd View Post
    Hello again Everyone,

    I had a question I was able to finally solve by myself but I was wondering if the following method is correct.

    The problem is, cot\theta = \frac{\sqrt{3}}{3}

    How I solved it...
    tan \theta = \frac{1}{cot \theta}

    tan \theta = \frac{1}{\frac{\sqrt3}{3}} Which equals 1.732

    Since I knew the question asked about special triangles I did the following
    tan(30) = 0.57735
    tan(45) = 1
    tan(60) = 1.73205
    Therefore \theta = 60

    Is this an acceptable method or the "right" method?
    It is definitely a method. I wouldn't necessarily say it is the most direct or elegant method.

    \displaystyle \begin{align*} \cot{\theta} &= \frac{\sqrt{3}}{3} \\ \cot{\theta} &= \frac{1}{\sqrt{3}} \\ \tan{\theta} &= \sqrt{3} \\ \theta &= 60^{\circ} + 180^{\circ} n \textrm{ where } n \in \mathbf{Z} \end{align*}
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  3. #3
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    Re: Question with special angles and Trig Identities

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