Thread: Help needed with some circle (arcs, tangents, etc) relations

1. Help needed with some circle (arcs, tangents, etc) relations

Sorry about this, I know people do not like solving my home work, but I would not have come here if I hadn't already spent an hour and a half searching on Google. Any way, I have a home work assignment and related test that involves the relations between circles, arcs, cords, tangent lines, secants, angles, etc. Since its too hard to describe as its a really complicated figure, I'll put a link to a Flickr page with a scanned version of the problem. If the photo is too small, hold ctrl and zoom in with the mouse wheel. Thanks guys.
001 on Flickr - Photo Sharing!

2. you should know the following

the radius which intersect with the tangent point is a perpendicular to the tangent line from that point

I will find as much as I can
so

<ACO = 90
<3 = 180 - <ACO = 90
<ABO ( <5) = 90
<COB = 360 - <ACO - <ABO - <CAB = 360 - 90-90-100 = 80
<2 = 180 - <COB = 100
<AOC = 180 - 70 = 110

in triangle DOE OD = OE two radius so <4 = <OED
so

2<4 + 70 = 180 , <4 = 55

<CED = 90 angle on the diameter

<ECD + <4 + 90 = 180 , <ECD = 45

in triangle BOD we have BO = DO why ?? so

<OBD = < ODB = (180 - <2)/2 = 40

<DCE = 1/2 <DOE center angle with circum angle at the same arc so

<DCE = 35

call the point of the intersection between CD and BF "x"

3. Thanks, but I'm pretty sure O is not the center. I don't remember how, but because then AOB would be 100 degrees, if you look at it that would not work based on the properties I do know. I think the properties we're supposed to be using are diffirent. Our teacher had us write down 4 relations, but I lost them. The only one I remember is that the segments of externally secant lines through the same point cut by the perimeter are proportional.

4. Originally Posted by chabos
Thanks, but I'm pretty sure O is not the center.
Given: $\displaystyle \odot O$

That pretty much tells you the center of the circle is O. And if you lost your notes, try reading the textbook.

5. Yeah, you're right, sorry. I was thinking in parallelograms, so I thought all angles would have to be 90 of the kite. About the theorems, they're not really in the book. I finally found them, they are hidden within the problems themselves as proofs. Its stuff like the angle of the intersection of two tangent lines is 1/2 the major arc minus the minor.

6. Here are the answers I got.

Angle 1: 80 degrees

This angle is formed by two tangents.

If two tangents intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the difference of the measures of the two intercepted arcs. 1/2(260-100)

Angle 2: 80 degrees

Angle AOB = 100 degrees because it is a central angle equal in measure to its intercepted arc.
Angle 2 is supplementary to angle AOB because they make up a linear pair on diameter DC.

Angle 3: 90 degrees

A radius OC is perpendicular to a tangent (AC) at the point of tangency C.

Angle 4 and Angle 5: 55 degrees

Triangle DOE is isosceles with a vertex angle of 70 degrees.

Angle 6: 90 degrees

Same reason as Angle 3

Angle 7: 65 degrees

Angle 7 can be found by determining arcs BD and CF. Angle 7 is one-half the sum of these two arcs. 1/2(80 + 50)

Arc BD = 80 because its the same measure as the central angle 2 which we found above.

Arc DE is 70 - same as the central angle which was given.

Arc CF can now be found by subtracting all the known arcs from 360. Arc CF is 50 degrees

Angle 8: 35 degrees

Angle 8 is one-half the measure of its intercepted arc DE.

Angle 9: ?? I don't see an angle 9 in the picture!

Arc BD: 80 degrees

Same as the central angle that intercepted it.

Arc BDE: 290 degrees

Everything except the arc DE.