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Math Help - chords

  1. #1
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    chords

    Hello everyone,

    Could someone please show me how to do this one?

    FInd the chord(120 degrees) when the radius of the triangle is 1 and then when the radius is 3438. I know that the answer is the square root of 3 for the first one, but I'm not sure how to go about solving it.

    My attempt:

    I know that the angles would have to be 120, 30, 30, but I'm not sure what to do next.

    Thank you very much
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  2. #2
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    Quote Originally Posted by Chocolatelover2 View Post
    Hello everyone,

    Could someone please show me how to do this one?

    FInd the chord(120 degrees) when the radius of the triangle is 1 and then when the radius is 3438. I know that the answer is the square root of 3 for the first one, but I'm not sure how to go about solving it.

    My attempt:

    I know that the angles would have to be 120, 30, 30, but I'm not sure what to do next.

    Thank you very much
    The "radius of the triangle"? Do you mean that the triangle is inscribed in a circle of radius 1? If so, then since 120= 360/3, you have an equilateral triangle. Let x be the length of the chord (one side of the triangle) and let r be the radius of the circle. Dropping a perpendicular to one side from the opposite angle divides the triangle into two right triangle having hypotenuse of length x and one leg of length x/2. By the Pythagorean theorem, the other leg, an altitude of the triangle, has length \frac{\sqrt{3}}{2}x.

    Now draw a line from the center of the circle to one of the other two vertices of the triangle. That, together with the altitude just drawn, gives a right triangle with hypotenuse r (the radius of the circle) and one leg of length x/2. The other leg, along the altitude, again by the Pythagorean theorem, has length \sqrt{r^2- \frac{x^2}{4}}. But that altitude is that leg plus a radius: \frac{\sqrt{3}}{2}x= r+ \sqrt{r^2- \frac{x^2}{4}} so that \frac{\sqrt{3}}{2}x- r= \sqrt{r^2- \frac{x^2}{4}}.

    Squaring both sides, \frac{3}{4}x^2- \sqrt{3}xr+ r^2= r^2- \frac{x^2}{4}. Now the " r^2" terms cancel giving \frac{3}{4}x^2+ \frac{1}{4}x^2= x^2= \sqrt{3}xr. Dividing both sides by x, x= \sqrt{3}r.

    Now take r= 1 and r= 3438 to answer your questions.
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  3. #3
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    Thank you very much
    Last edited by Chocolatelover2; March 29th 2009 at 12:12 PM.
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