# Angles - Circle Properties

• Jan 7th 2010, 02:17 AM
Punch
Angles - Circle Properties
http://i952.photobucket.com/albums/a...g/P1070411.jpg

In the diagram, TP and TQ are tangents to the circle at the point P and Q respectively. TQY is a straight line. Given that ∠RPQ = 26°, ∠SQY = 42° and ∠RXQ = 101°, calculate the following angles.
(a) ∠QRS
(b) ∠RQP
(c) ∠RQS
(d) ∠PTQ
• Jan 7th 2010, 02:49 AM
Hello Punch
Quote:

Originally Posted by Punch

In the diagram, TP and TQ are tangents to the circle at the point P and Q respectively. TQY is a straight line. Given that ∠RPQ = 26°, ∠SQY = 42° and ∠RXQ = 101°, calculate the following angles.
(a) ∠QRS
(b) ∠RQP
(c) ∠RQS
(d) ∠PTQ

Do you know the Alternate Segment Theorem? This gives:

(a) $\angle QRS = \angle SQY = 42^o$

(b) $\angle RQP$ can then be calculated from the angles of $\triangle RQX$.

(c) Using the alternate segment theorem again, $\angle RQT = \angle RPQ = 26^o$. So you can now find $\angle RQS$ from the straight line $TQY$.

(d) $\angle TQP = \angle TQR + \angle RQX=\angle TPQ$ (since $\triangle TPQ$ is isosceles). So you can now work out $\angle PTQ$.

Can you fill in the gaps?

• Jan 7th 2010, 06:48 AM
Amer
as I can see you find most angles but you can find all of them by finding QPS

there is a theorem said that the angle between the tangent of the cirlce and the chord equal to the angle on that chord so

QPS=SQY = 42

I think now you can find all

you can use this to find RQT and RPT

RQT = QSR
RPT = RSP
• Jan 7th 2010, 07:22 AM
Shanks
Yeah, Amer is right!
When I was studying geometry in the plane, I saw the circle to be the most Unbelievable math Object.(Rofl)