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Math Help - triangle inscribed in a semicircle

  1. #1
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    triangle inscribed in a semicircle

    Can someone tell me why a triangle inscribed in a semicircle is always a right triangle?

    I am in Trig currently and someone told me I had to use the laws of cosine. Which I am familiar with, I just don't know how to prove it.

    Thank you
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  2. #2
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    Quote Originally Posted by zodiacbrave View Post
    Can someone tell me why a triangle inscribed in a semicircle is always a right triangle?

    I am in Trig currently and someone told me I had to use the laws of cosine. Which I am familiar with, I just don't know how to prove it.

    Thank you
    This is a special case of Inscribed and Central Angles in a Circle from Interactive Mathematics Miscellany and Puzzles

    When you read the proof keep in mind that your central angle is 180.
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  3. #3
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    Quote Originally Posted by zodiacbrave View Post
    Can someone tell me why a triangle inscribed in a semicircle is always a right triangle?

    I am in Trig currently and someone told me I had to use the laws of cosine. Which I am familiar with, I just don't know how to prove it.

    Thank you
    If it should be proven by the Law of Cosines, then,
    let yhe two legs of the inscribed triangle be x and y
    the hypotenuse is diameter d.
    we assume the angle between x and r is 90 degrees.....this angle is opposite d.

    By Law of Cosines,
    d^2 = x^2 +y^2 -2(x)(y)cos(90deg)
    d^2 = x^2 +y^2 -2xy(0)
    d^2 = x^2 +y^2 -------------this is the Pythagorean Theorem, applicable to right triangles only. So the inscribed triangle is really a right triangle, whatever the lengths of x and y are.
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  4. #4
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    Sorry, your Web browser is not Java compatible. This prototype will not be of much interest to you.


    It sometimes helps to see a dynamic diagram.

    There should be a java applet to the right of this text (it might take a second to load)


    You can move the red dots around.


    Notice that angle C is always a right angle.




    On a side note to others: How many knew that you could type in a java applet? I used to use it all the time, but I haven't been here in like a year.
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  5. #5
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    Hello, zodiacbrave!

    Well, here's one way . . . my way.


    Why is a triangle inscribed in a semicircle is always a right triangle?
    We have a circle with center O, diameter AB, and radii  OA = OB = OC = r.
    Draw chords AC = x and BC = y.
    Let \angle AOC = \theta,\;\;\angle BOC = \theta'
    Code:
                  * * *
              *           *   C
            *               o
           *           r  *  *
                        *
          *         θ * θ'    *
        A o - - - - o - - - - o B
          *    r    O     r   *
    
           *                 *
            *               *
              *           *
                  * * *

    Use the Law of Cosines in \Delta AOC\!:\;\;x^2 \:=\:r^2 + r^2 -2(r)(r)\cos\theta
    . . We have: . x^2\;=\;2r^2 - 2r^2\cos\theta .[1]

    Use the Law of Cosines in \Delta BOC\!:\;\;y^2 \;=\;r^2 + r^2 - 2(r)(r)\cos\theta'
    . . We have: . y^2 \;=\;2r^2 - 2r^2\cos\theta' .[2]

    Since \theta + \theta' \:=\:180^o\quad\Rightarrow\quad \theta' \:=\:180^o - \theta
    . . \cos\theta' \:=\:\cos(180^o-\theta) \:=\:-\cos\theta

    Substitute into [2]: . y^2 \;=\;2r^2 - 2r^2(\text{-}\cos\theta)
    . . and we have: . y^2 \;=\;2r^2 + 2r^2\cos\theta .[3]

    Add [1] and [3]: . x^2 + y^2 \;=\;(2r^2 - 2r^2\cos\theta) +(2r^2+2r^2\cos\theta)

    . . x^2 +y^2 \:=\:4r^2 \quad\Rightarrow\quad x^2 + y^2 \:=\:(2r)^2 . . . Pythagorus!


    Therefore, \Delta ABC is a right triangle with \angle C = 90^o.

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