# Thread: Angles - Circle Properties

1. ## Angles - Circle Properties

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

2. Hello Punch
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?

3. 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

4. Yeah, Amer is right!
When I was studying geometry in the plane, I saw the circle to be the most Unbelievable math Object.