# No idea...transversal

• Jul 26th 2009, 08:45 AM
A Beautiful Mind
No idea...transversal
http://img187.imageshack.us/img187/4103/yesere.png

Confused here. Would CQF be 110 as well? Thus meaning that on the other side there exists 70 degrees?
• Jul 26th 2009, 09:04 AM
red_dog
Quote:

Originally Posted by A Beautiful Mind

Confused here. Would CQF be 110 as well? Thus meaning that on the other side there exists 70 degrees?

No. \$\displaystyle \widehat{DQF}=\widehat{APE}=110\$
• Jul 26th 2009, 10:41 PM
Conorsmom
Quote:

Originally Posted by red_dog
No. \$\displaystyle \widehat{DQF}=\widehat{APE}=110\$

Yet another way to look at it : If AB is ll to CD, then angles APE and CQF are congruent, being 180 degrees as given.

This means that Angle CQF must be 70 degrees by the supplementary angle theorem.

This leaves the measurement of FQD to be 110 degrees, and being bisected by ray QR, means that the measurement of angle FQR is 55 degrees, 1/2 of FQD.

Now add m <FQR and m<CQF and you get A) 125 degrees.