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Math Help - Circle in Triangle

  1. #1
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    Circle in Triangle

    Hi I have attached the question as a word file. I have worked through it, and need help with the last bit.

    I do not understand how to show the relationship between A and B in ii)

    he first time I did the question, I tried to use the relationship to evaluate the areas asked for next, which ended up being confusing.

    However, this time, I just worked with the kite and sector of the circle to find A. I didn't use B at all.

    Since B was given right before the question to work out the final 2 Areas was asked, I am presuming that it can be used to work out the answers, or else it would have no reason of being added.

    So basically, I dont understand the relationship, and also, can anyone show how they would do the last part using this relationship between area A and B?

    Thanks
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  2. #2
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    Hello Aquafina
    Quote Originally Posted by Aquafina View Post
    ... So basically, I dont understand the relationship, and also, can anyone show how they would do the last part using this relationship between area A and B?

    Thanks
    The function A(\theta) gives the area between any two tangents and the circle expressed as a function of \theta, the angle between the tangents.

    So if we denote the angle between the tangents from R to the circle by \phi, then the area between these tangents and the circle is A(\phi). But this area is given by B(\theta). Hence A(\phi) = B(\theta).

    But, from \triangle ORP, \theta = \frac{\pi}{2}-\phi. Therefore A(\phi) = B(\frac{\pi}{2}-\phi). We can replace \phi by any other parameter; in particular by \theta. So A(\theta) = B(\frac{\pi}{2}-\theta).

    For part (iii) I should do as you have done. The calculation is very straightforward for \theta = \frac{\pi}{3}. Why complicate it?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello AquafinaThe function A(\theta) gives the area between any two tangents and the circle expressed as a function of \theta, the angle between the tangents.

    So if we denote the angle between the tangents from R to the circle by \phi, then the area between these tangents and the circle is A(\phi). But this area is given by B(\theta). Hence A(\phi) = B(\theta).

    But, from \triangle ORP, \theta = \frac{\pi}{2}-\phi. Therefore A(\phi) = B(\frac{\pi}{2}-\phi). We can replace \phi by any other parameter; in particular by \theta. So A(\theta) = B(\frac{\pi}{2}-\theta).

    For part (iii) I should do as you have done. The calculation is very straightforward for \theta = \frac{\pi}{3}. Why complicate it?

    Grandad
    Thanks Grandad. I was told I had to mention something like:

    That the reflections would not change the areas and reverses the angles...

    Also, I wanted to try the relationship with B to work out the Area of A just to experiment around. But this didnt work out.

    I did

    A(pi/3) = B(pi/6) = 2*AreaQCR - Area segment
    = root3 - 5pi/12

    rather than the pi/3...

    I get pi/3 if I take the angle ORP to be pi/3, not pi/6
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  4. #4
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    Hello Aquafina
    Quote Originally Posted by Aquafina View Post
    Thanks Grandad. I was told I had to mention something like:

    That the reflections would not change the areas and reverses the angles...

    Also, I wanted to try the relationship with B to work out the Area of A just to experiment around. But this didnt work out.

    I did

    A(pi/3) = B(pi/6) = 2*AreaQCR - Area segment
    = root3 - 5pi/12

    rather than the pi/3...

    I get pi/3 if I take the angle ORP to be pi/3, not pi/6
    I'm not quite sure what you mean here. The straightforward proof that A(\frac{\pi}{3})= \sqrt3-\frac{\pi}{3} is as follows:

    If C is the centre of the circle, then \angle CPQ = \frac{\pi}{6}

    \Rightarrow PQ = \frac{1}{\tan(\tfrac{\pi}{6})} = \sqrt3

    \Rightarrow area of \triangle CPQ = \tfrac12\cdot1\cdot\sqrt3 = \frac{\sqrt3}{2}

    \Rightarrow area of kite = 2\times\triangle CPQ = \sqrt3

    The angle at the centre of the circle is \pi - \theta = \frac{2\pi}{3}

    \Rightarrow area of sector = \tfrac12\times1^2\times\frac{2\pi}{3}= \frac{\pi}{3}

    \Rightarrow A(\frac{\pi}{3}) = area of kite - area of sector = \sqrt3 - \frac{\pi}{3}

    Grandad
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