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Math Help - Finding the value of the angle theta in radians, knowing sec theta and cot theta

  1. #1
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    Finding the value of the angle theta in radians, knowing sec theta and cot theta

    Hi,

    The question I have is:

    Given that \sec \theta = -2, \cot \theta = -1/3\sqrt{3} and -\pi < 0 < \pi, find the exact value of the angle \theta in radians. Justify your answer.

    Is the following along the right lines?

    \sec\theta=1/cos\theta therefore \cos\theta=1/\sec \theta=-1/2
    \cot\theta=1/\tan\theta therefore \tan\theta=1/\cot\theta=-3 \sqrt{3}/3

    \sin\theta =\tan\theta\cos\theta=-3\sqrt{3}/3x-1/2=1/2sqrt{3}

    Since both \tan\theta and \cos\theta are negative and lie in the range -\pi < 0 < \pi, then \theta must lie in the 2nd quadrant so is positive.

    Thanks for your help
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  2. #2
    MHF Contributor

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    Quote Originally Posted by cozza View Post
    Hi,

    The question I have is:

    Given that \sec \theta = -2, \cot \theta = -1/3\sqrt{3} and -\pi < 0 < \pi, find the exact value of the angle \theta in radians. Justify your answer.

    Is the following along the right lines?

    \sec\theta=1/cos\theta therefore \cos\theta=1/\sec \theta=-1/2
    \cot\theta=1/\tan\theta therefore \tan\theta=1/\cot\theta=-3 \sqrt{3}/3

    \sin\theta =\tan\theta\cos\theta=-3\sqrt{3}/3x-1/2=1/2sqrt{3}

    Since both \tan\theta and \cos\theta are negative and lie in the range -\pi < 0 < \pi, then \theta must lie in the 2nd quadrant so is positive.

    Thanks for your help
    Good. Now you also need to know some basic angles. All three angles in an equilateral triangle have measure \pi/3 radians. If you drop a perpendicular from one vertex to the opposite side, you divide the triangle into two congruent right triangle with angles of \pi/3 and [/tex]\pi/6[/tex]. If you take the equilateral triangle to have sides of length 1, the hypotenuse of the right triangles is 1 and the leg opposite the \pi/6 angle is 1/2.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Good. Now you also need to know some basic angles. All three angles in an equilateral triangle have measure \pi/3 radians. If you drop a perpendicular from one vertex to the opposite side, you divide the triangle into two congruent right triangle with angles of \pi/3 and [/tex]\pi/6[/tex]. If you take the equilateral triangle to have sides of length 1, the hypotenuse of the right triangles is 1 and the leg opposite the \pi/6 angle is 1/2.

    Thank you for such a quick reply! So does the angle \theta=1/2sqrt{3}? Can I then justify this answer by explaining what you have written? Why do I need to know the lenths of the sides in the right angle traingle to find the exact value of the angle \theta in rtadians? Sorry if this seems like a really stupid question.
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