question on properties of circle 6

**ukorov** · November 8th 2009, 04:31 AM

Referring to the attached pic:
A quadrilateral ACQR is inscribed in a circle. Diagonals are drawn to intersect at T such that angle $ATR = 76^o$
Both S, P and B are points outside the circle.
AC and RQ are extended to meet at P such that angle $APR = 30^o$
CQ and AR are extended to meet at B such that angle $ABC = 46^o$
QC is extended to S, forming triangle SAC where SC = AC

Find the angle ASC.

**red_dog** · November 8th 2009, 06:06 AM

$\widehat{ART}=\frac{m(arcAR)+m(arcCQ)}{2}=76$

$\widehat{APR}=\frac{m(arcAR)-m(arcCQ)}{2}=30$

Then $m(arcAR)=106, \ m(arcCQ)=46$

$\widehat{ATC}=\frac{m(arcAC)+m(arcQR)}{2}=104$

$\widehat{ABC}=\frac{m(arcAC)-m(arcQR)}{2}=46$

Then $m(arcAC)=150, \ m(arcQR)=54$

Now, $\widehat{ACQ}=\frac{m(arcQR)+m(arcAR)}{2}$

Can you continue?

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