MisterHewson
New Member
- Joined
- Nov 12, 2012
- Messages
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- Male
- HSC
- 2013
Angle between a chord and a tangent is equal to the angle in the alternate segment?This is how I did it, which I'm pretty sure is right:
Angle between a chord and a tangent is equal to the angle in the alternate segment?This is how I did it, which I'm pretty sure is right:
This is pretty much what I did. However, I did do the beta part. I just wrote angle QTX = angle PTY = alpha, therefore P, T and Q are collinear because vertically opposite angles are equal. Would this be accepted?This is how I did it, which I'm pretty sure is right:
No. [Assuming you said vertically opposite angles first] iBibah got angle QTX as alpha from a different argument. Cannot say vertically opposite because there is no guarantee QTP is straight UNLESS you used a different argument of finding alpha.This is pretty much what I did. However, I did do the beta part. I just wrote angle QTX = angle PTY = alpha, therefore P, T and Q are collinear because vertically opposite angles are equal. Would this be accepted?
I didn't use vertically opposite angles first. To get the alphas, I used the alternate segement theroem, then I conclude that P, T and Q are collinear because the vertically opposite angles are equal.No. [Assuming you said vertically opposite angles first] iBibah got angle QTX as alpha from a different argument. Cannot say vertically opposite because there is no guarantee QTP is straight UNLESS you used a different argument of finding alpha.
That's the way I did it :SI sat my HSC last year but looking at this question would this be valid:
let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)
Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)
Therefore angle QTP = x + y + 180 - (x +y)
= 180
Thus Q,T,P are collinear
+1 It's valid because you prove they're equal angles and THEN you state well since they're equal, they must be vertically opposite. Hence straight line.I didn't use vertically opposite angles first. To get the alphas, I used the alternate segement theroem, then I conclude that P, T and Q are collinear because the vertically opposite angles are equal.
Not valid, because you assumed that QBP was a straight line, which was what the question was asking you to prove.I sat my HSC last year but looking at this question would this be valid:
let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)
Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)
Therefore angle QTP = x + y + 180 - (x +y)
= 180
Thus Q,T,P are collinear
I would think it would be 1/3 though I am not familiar with the marking guidelines. But 1/3 makes sense.Do you still think that deserves 1/3 since the first statement is correct in completing the proof?
That's assuming the result, so no.I sat my HSC last year but looking at this question would this be valid:
let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)
Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)
Therefore angle QTP = x + y + 180 - (x +y)
= 180
Thus Q,T,P are collinear
Maybe 1/3 for using the alternate angles between parallel lines on the line that was given to you as straight,Do you still think that deserves 1/3 since the first statement is correct in completing the proof?