# Geometry Circles

• Apr 28th 2009, 11:59 AM
Geometric
Geometry Circles
Hello,
I need help urgently on how to solve these problems I am extremely confused :
(I think that 1 a) 100 but I can't figure out how to get the other ones.)

http://img528.imageshack.us/img528/6...10f731f2d8.jpghttp://img528.imageshack.us/img528/3...4f4c21ab5e.jpg
• Apr 28th 2009, 01:33 PM
masters
Quote:

Originally Posted by Geometric
Hello,
I need help urgently on how to solve these problems I am extremely confused :
(I think that 1 a) 100 but I can't figure out how to get the other ones.)

http://img528.imageshack.us/img528/6...10f731f2d8.jpghttp://img528.imageshack.us/img528/3...4f4c21ab5e.jpg

Hi geometric,

There is quite a lot of explanation that should accompany these answers. I hope you've studied your circle theorems.

(a) \$\displaystyle m \widehat{BC}=100^{\circ}\$ because you were given that \$\displaystyle m \angle CAB = 50^{\circ}\$, which is half of the intercepted arc BC.

(b) \$\displaystyle m \angle EBA =40^{\circ}\$ because you can find the measure of arc AE by subtracting the other known arcs from 360. Arc AE measures 80 degrees. That makes the inscribed angle = 40.

(c) \$\displaystyle m \angle ADE = 90^{\circ}\$ because the angle formed by the intersection of two chords is one-half the sum of the measures of the two subtended arcs. Arc AE = 80 degrees and arc BC = 100 degrees. Their sum is 180, and half that is 90 degrees.

(d) \$\displaystyle m \angle F=35^{\circ}\$ because the exterior angle formed by two secants is equal to one-half the difference of its intercepted arcs. Thus angle F is 1/2(150 - 80) or 35 degrees.

(e) \$\displaystyle m \angle FAC = 55^{\circ}\$ since it is in the right triangle FAD and is complementary to angle F.

Your second question requires even more explanation. I suggest you post it separately and tell us how far along you have gotten. Be specific about what you don't understand.

Study your circle theorems (all about central angles, inscribed angles, tangents, secants, chords, arcs, etc.).
• Apr 28th 2009, 01:41 PM
masters
Quote:

Originally Posted by Geometric
Hello,
I need help urgently on how to solve these problems I am extremely confused :
(I think that 1 a) 100 but I can't figure out how to get the other ones.)

http://img528.imageshack.us/img528/3...4f4c21ab5e.jpg

Geometric,

In your second one, you can find the measures of the angles of the inscribed triangle this way:

\$\displaystyle m \angle BAC:m \angle CBA:m \angle ACB=2:3:4\$

Thus, \$\displaystyle 2x+3x+4x=180\$

Solve for x, then substitute to find the measures of each angle.

\$\displaystyle m \angle BAC=2x\$

\$\displaystyle m \angle CBA=3x\$

\$\displaystyle m \angle ACB=4x\$

• Apr 28th 2009, 01:46 PM
Geometric
Thank you very much for the help with #1. After re-reading and reading #1's explanation I came up with these answers for #2 :

a) 40
b) 80
c) 80 or 100? ( this is the one that confused me)
d)60
e) 20
• Apr 28th 2009, 02:08 PM
masters
Quote:

Originally Posted by Geometric
Thank you very much for the help with #1. After re-reading and reading #1's explanation I came up with these answers for #2 :

a) 40
b) 80
c) 80 or 100? ( this is the one that confused me)
d)60
e) 20

Geometric,