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Prove ∠XOM = θ

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data Q7 part a on one of the attached pictures

    2. Relevant equations

    Trigonometric identities

    3. The attempt at a solution
    See attached pages



    Please help me I've spent onwards of 4 hours trying to figure this out and I can't get anywhere at all
     

    Attached Files:

  2. jcsd
  3. Apr 23, 2016 #2
    I have doubt about the problem. What is the guarantee that the line segment ZM passes through O the center of the circle. Does it mean that this is also given as input data.
     
  4. Apr 23, 2016 #3
    ∠XOY=2θ(angle at center twice angle at circumference)
    XO=OY=raidius
    ∠XOM=∠YOM(angle at same segment)
    2∠XOM=2θ
    ∠XOM=θ
     
  5. Apr 23, 2016 #4
    All I know is what I was given in the question. But I agree the question is presumptuous; it's assuming h≥r
     
  6. Apr 23, 2016 #5
    How do you know angle XOY is double angle XZY? Is it just a rule that I don't know about?
     
  7. Apr 23, 2016 #6
     
  8. Apr 23, 2016 #7
    Thank you so much that has helped heaps
     
  9. Apr 23, 2016 #8
    The rest of the problem is quite simple. We need to solve the problem as mentioned in the diagram h = ZO +OM = r + rcos θ or
    h/r = cos θ + 1. You can find dh/dθ and put r = 3 and θ = π/6 to complete the answer. The diiagram given in the book is a special case of the inscribed triangle where ZM passes through O.
     
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