# Circles and Tangents.

• Jul 7th 2005, 12:49 AM
cornoth
Circles and Tangents.
Hi guys again. I drew a diagram for this question, and I get 75 degrees for ABQ, after some shuffling of angles/ etc. I used Euclid III.32, to get to the solution as the tangent and the radius is 90 degrees, but there since PBT is 75 degrees I thought that ABQ is also 75 degrees.

However my diagram is a mess of lines and I still am not sure if the answer is 15 degrees? or 75 or something else. Although on my diagram it looks like 15, but you can never rely on diagrams :)

I was hoping for a clear explanation. Thanks to all!

Two circles C1 and C2 meet at the points A and B. P is a point on the circumference of C1 such that PA is tangent to C2 at A. Q is a point on the circumference of C2 outside C1. QA is produced to meet C1 again at T
so that anglePBT = 75 degrees. Find the size of angleABQ.
• Jul 8th 2005, 09:12 PM
ticbol
Clarification
This question bugs me everytime I look at it. I think there is lacking information. As posted, I cannot sketch a correct figure based on the given information.
Without a correct figure, I cannot analyze the problem correctly.

If only circles C1 and C2 equal....

If they are, then, yes, angle ABQ is 75 degrees also, and I can show that.

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Did you post the question in full?
• Jul 9th 2005, 04:36 AM
ticbol
Zeez, why didn't I see this before?

Let me take back my wish that C1 be same size as C2. I just found out that whatever are the sizes of circles C1 and C2, the angle ABQ is always 75 degrees.
There is no lacking information then from the question as posted.

Let us sketch the figure.

Draw an x,y coordinate axes setup. Let the origin (0,0) be point A.
Draw circle C2, of any size, such that its topmost is at point A. So the center of C2 is along the y-axis, below the origin. Let us call the bottom most point as point D. AD then is the vertical diameter of circle C2.
Draw cirle C1, of any size also, to the left of C2, such that C1 and C2 intersect at point A and point B. The center of C1 needs not be on the x-axis.
But line PA is on the x-axis because PA is tangent to C2 at point A. Draw line PA.
Draw line AB.
Draw line PB.
Draw line BT, such that angle PBT is about 75 degrees. (About 75 degrees is for sketching only. The given fact is angle PBT is really 75 degrees.)
From point T on the circumference of C1, draw straight line TAQ, point Q being on the circumference of circle C2.
Then, draw line BQ.

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In circle C1:

Angle(PBT) is an inscribed angle in C1, and is subtended by minor arc PT.
Since angle(PBT) is 75 degrees, then arc PT is 2*75 = 150 degrees.

>>>That is, the measure of an inscribed angle is half the measure of the arc subtended by the angle. Or, the subtended arc is twice the measure of the inscribed angle.

Angle(PAT) is also an inscribed angle in C1, and it is also subtended by arc PT.
Hence, angle(PAT) = 150/2 = 75 degrees also.

Let us call the other intersection point of the y-axis and C1 as point E.
Angle(PAE) is the 2nd quadrant, and angle(PAE) = 90 degrees.
So, angle(TAE) = 90 -75 = 15 degrees.

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In circle C2:

Angle(DAQ) and angle(TAE) are vertical angles. Vertical angles are equal in measure, hence, angle(DAQ) = 15 degrees too.

Angle(DAQ) is an inscribed angle in C2, and it is subtended by minor arc DQ.
So, arc DQ = 2(15) = 30 degrees

Arc DQA is a semicircle, so arc DQA = 360/2 = 180 degrees.
Hence, minor arc QA = 180 -30 = 150 degrees

Angle ABQ is an inscribed angle in C2, and it is subtended by minor arc QA.
Therefore, angle(ABQ) = (1/2)(150) = 75 degrees ------answer.
• Jul 12th 2005, 01:15 AM
cornoth
Thanks ticbol, once again excellent and detailed. :o