Question

Hey guys, I have several problems that look like this, could you show me how to do this one so I can knock them all out? They don't seem hard, I just need to know how to do them, thanks!

Answer 1

In whatever format you like best :)

Answer 2

I know only that AD=DC
The tangents are perpendicular to the radius at the point A and C.
The angle at the center of the circle formed by the radius from A and from C forms an angle twice the size of ABC.

Answer 3

is there any way we can figure out the measure of arc ABC? Because if I could figure that out I think I could figure out what angle D is.

Answer 4

Well the angle at the center formed by the radii from A and C would be 100 degrees. Twice angle B. The arc is the same as the angle. The arc would be 100.

Answer 5

How do we know that A and C are radii from the center?

Answer 6

No. AD is tangent, CD is tangent. You can draw a radius from center to point A, likewise you can draw a radius from center to point D.

Answer 7

So on problems like these that look exactly the same, only with angles with different values, I can just times B by two to get the arc for ABC?

Answer 8

From what I can deduce, I just looked this up because of your questions. It would be more specifically the minor arc.

Answer 9

oh so AC would be 100?

Answer 10

Let's call the center of the circle E. The angle AEC would be 100 degrees.

Answer 11

okay... any ideas on how to proceed :(

Answer 12

I don't know the properties of the circle involved, so I would have to figure it out long-hand. I can figure out the radius\[\theta =arc/r\]I can draw a bisector from center E to D. ADE is a right triangle with right angle at A and angle 50 degrees at E. With that I can find D.

Answer 13

so it would be 40 times 2 which is 80 as final answer?

Answer 14

I don't know where you got those numbers. Please explain.

Answer 15

Oh, I understand. You surmised the third angle must be 40. Yeah, I agree with you.