# Angles - Circle Properties

• Jan 7th 2010, 01:17 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, 01:49 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) \$\displaystyle \angle QRS = \angle SQY = 42^o\$

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

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

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

Can you fill in the gaps?

• Jan 7th 2010, 05:48 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, 06:22 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)