Need help with Algebra 2 problem?

Uh-oh! Two different Answers?

I wish I could draw this out....but    basically   40 + (2x +20) + (4x + 40) = 360 degrees

solve for x        x = 43 2/3 degrees

If you look at the vertex of the angle you have one arm that has rotated 40 degrees.....rotate that arm another  2x+20 degrees   then another 4x + 40 degrees for a complete circle (360 degrees)

So  40 + (2x+20) + (4x+40) = 360

~jc

The inscribed angle in the circle measures 1/2 of its intercepted arc.......so z°   = 80°

Then.....the other two arcs must comprise 360 - 80  =  280°  of arc.......therefore

(2x + 20) + (4x + 40)  = 280   simplify

6x + 60  =  280     subtract 60 from both sides

6x = 220     divide both sides by 6

x = [36 2/3]°

(C) 

Hey Chris.....   Help me....  How do you know that the inscribed angle  measures 1/2 of it's intercepted arc??

Thanx

~jc

Here you go, jc   http://www.regentsprep.org/Regents/math/geometry/GP15/CircleAngles.htm

Look at the second property......

Chris to the rescue!!

Thanks so much, you guys are awesome :)

Hey Chris.....   Help me....  How do you know that the inscribed angle  measures 1/2 of it's intercepted arc??

Thanx

I found this: 
Inscribed Angle =  1/2   Intercepted Arc

 Does this ALWAYS apply even if the legs are not equal length??

NICE WORK CPhill  !!!!

~jc

Yep....it always applies......the length of the chords comprising the angle's sides don't matter....

All right!   Learned something NEW today.  There's a new shiny spot where some rust used to be.  (Probably just forgot it over the years)..

If you are able , you can delete my answer to avoid confusion....

~jc

Sup jc?

Hello Jc !

Hey Coldplay....howz NOLA ?

Hi Hayley...

Just hanging out today...preparing/packing my bike to go riding to a hot springs in the National Forest up by Jackson Hole WY tomorrow....

Nice! I'm half asleep right now, So.. I might say gofy things by mistake..

So..

Heads up!