Would appreciate help with the following (both links are for the same Q) This was from 2009HSC last Q but I don't understand the provided solution.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21984023_1334137693378417_1111165516_n.png?oh=065a246eb62d7e0b274077d3b0122249&oe=59C9A7BE
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21984439_1334137783378408_2124958823_n.png?oh=8e2351abe5ad2855f9e7f499e35c6b88&oe=59C9BFCF
Parts (c). Thanks in advance!
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The BOSTES answers probably aren't the best, but ultimately this question then relies on intuition. I was hoping for something better but I couldn't find any.
Firstly, consider the case where P is to the left of T, as shown in the diagram. We will
gradually move the point P towards the point T. Once the points P and T coincide, we no longer have \(\theta < \phi\), but rather \(\theta = \phi).
This happens because we must keep in mind that Q, P and S are collinear. As we gradually move P towards T, we are moving
S towards T as well.This is why when P and T coincide, S coincides with it
as well, which then imply that the angles \( \theta \) and \( \phi \) coincide.
GeoGebra simulation attached. You will find that as P gets dragged away from T, \( \theta\) gets smaller. And when P gets dragged towards T, \( \theta\) gets larger.
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Now that's one half of the story. The other half is when P is to the right of T instead.
You can repeat the same thing we did earlier. In a similar way, if P is to the
right of T,
we still have \( \theta < \phi \). And as we move P to the left, gradually coinciding with T again, eventually we will have \(\theta = \phi\) once they coincide.
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From there on, we need to recollect what we know.
Remark: Perhaps the most curious thing about this is that \( \phi \) is a
fixed angle, which is what allows this argument to work. The reason for this is because of the circle geometry theorem "angles in the same segment are equal", aka. "angles standing on the same arc are equal". Here, the arc is just QR, and we're moving the angle along the major arc.
(Note, however, that phi doesn't necessarily have to be 28 degrees. That exact value was just a consequence of how I simulated it.)
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(May need to stare at the diagram really closely to see this.)