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Thread: proving ab subtends c at right angle

  1. #1
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    proving ab subtends c at right angle

    A circle, radius two units, center the origin , cuts the x-axis at A and B and cuts the positive y-axis at C. Prove that AB subtends a right angle at C.


    proving ab  subtends  c at right  angle-q1.jpg
    I have tried drawing it so i can understand the problem better in order to answer the question. but looking at the sketch. It creates an isosceles
    triangle not a right angle

    I have tried a different sketch;

    proving ab  subtends  c at right  angle-q2.jpg
    the problem with this sketch is that c is not the positive y-axis.

    any tips so i can figure it out
    thanks
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  2. #2
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    Re: proving ab subtends c at right angle

    A circle, radius two units, center the origin , cuts the x-axis at A and B and cuts the positive y-axis at C. Prove that AB subtends a right angle at C.
    What kind of triangles are BOC and AOC? How do you know?

    What is the measure of angle BCO? ACO? ... so, what is the measure of angle BCA?

    link fyi ...

    Angle inscribed in a semicircle - Math Open Reference

    proving ab  subtends  c at right  angle-inscribed_semi.jpg
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    Re: proving ab subtends c at right angle

    Quote Originally Posted by bigmansouf View Post
    A circle, radius two units, center the origin , cuts the x-axis at A and B and cuts the positive y-axis at C. Prove that AB subtends a right angle at C.
    You are being asked to prove analytically that any angle inscribed in a semi-circle is a right angle.
    The x-intercepts are $A: (-2,0)~\&~B: (2,0)$. Any point of the upper semi-circle is $C: (a,\sqrt{4-a^2}),~-2\le a\le 2$

    The slopes $S(\overline{AC})=\dfrac{\sqrt{4-a^2}}{a+2}~\&~S(\overline{BC})=\dfrac{\sqrt{4-a^2}}{a-2}$

    Can you show that $S(\overline{AC})\cdot S(\overline{BC})=-1~?$
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    Re: proving ab subtends c at right angle

    Quote Originally Posted by bigmansouf View Post
    I have tried drawing it so i can understand the problem better in order to answer the question. but looking at the sketch. It creates an isosceles triangle not a right angle
    Ever heard of a right-isosceles triangle (45-45-90)?
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