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Comparing angles opposite radius of a circle

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    In circle O , BC >CD
    Compare x & y (which is greater?)

    2. Relevant equations
    There are no eq. Rule: angle opposite larger side is larger

    3. The attempt at a solution
    In my view both triangles are isosceles triangle. So x & y should be equal because they are both opposite the radius
    But book says a different answer. Some body please guide me.

  2. jcsd
  3. Dec 4, 2016 #2
    Why ?
    Imagine a 6,8,10 right triangle and a equilateral triangle with side 6. Is in both cases angle opposite to side 6 same as 60° (use trigonometry if you don't have the time to construct the triangles).
    Last edited: Dec 4, 2016
  4. Dec 4, 2016 #3
    <Why ?>
    OC is the radius of circle.
    <6,8,10 right triangle>
    In this case, the angle opposite to 6 would be the smallest.
    <equilateral triangle>
    In this case angle should be 60 & sides must be equal not necessary 6.

  5. Dec 4, 2016 #4
    Yes, but does that make the angle x and y equal ?

    Yes you are correct.
    So what do we deduce from this ?
    Also compare ##\angle BOC##and ##\angle COD##, which is greater ?
    Last edited: Dec 4, 2016
  6. Dec 4, 2016 #5
    Since its given that BC is greater therefore, arc BC is greater so angle BOC> DOC. But I don't know about x & y? Only that they are opposite to radius so they must be equal.

  7. Dec 4, 2016 #6


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    Right. So can you write x in terms of angle BOC and y in terms of DOC?
  8. Dec 5, 2016 #7


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    There is no such rule. E.g. consider B starting near C and moving around to be almost opposite C. At first, x is nearly 90o; half way round it is 45o; when opposite it is 0.
    The angle subtended by a diameter is always a right angle. Maybe you are thinking of that.
  9. Dec 6, 2016 #8
    Thanks for your example. But i cant understand it. But i want to give you an example of isosceles triangle in which angles opposite to the equal sides are equal and the triangle formed by connecting two radii is an isosceles triangle.
    <So can you write x in terms of angle BOC and y in terms of DOC?>
    Based upon my understanding:
    x + <BOC +<BCO = 180
    or <BOC= 180 -2x
    <DOC = 180 -2y
    Kindly guide me.
  10. Dec 6, 2016 #9


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    So x=? and y=? (in terms of BOC and DOC). If x and y were equal, BOC and DOC would also be equal. But BOC>DOC, meaning x and y are not equal. Which one is greater then?
  11. Dec 6, 2016 #10
    $$x = {180- \angle BOC \over 2}$$
    $$y = {180- \angle DOC \over 2}$$

    Now which is bigger ?
  12. Dec 6, 2016 #11
    Your case is true for two angles opposite to equal sides in a triangle. Here we are talking about two different triangles with different bases. So your rule will not work here.
  13. Dec 6, 2016 #12
    From these equation it looks that y is bigger. But you have derived this eq based upon my work and i used the logic that ∠x is equal to ∠BCO because BOC is an isosceles triangle and ∠x is opposite to OC. Similarly ∠y is opposite to OC, so its astonishing that ∠x & ∠y are not equal?

  14. Dec 6, 2016 #13


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    Yes, BOC is isosceles, with OB=OC, so angle OBC=angle OCB.
    Likewise, angle OCD=angle ODC.
    But these are two different isosceles triangles; there is no rule that says they must have the same angles as each other, even if they have the same lengths on some sides.
    Simple experiment: draw a triangle with sides length 4, 4, 7 and another with sides length 4, 4, 1. Each has two equal angles, but that angle is very different in one from the other.
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