Question

The figure below shows a triangle with vertices A and B on a circle and vertex C outside it. Side AC is tangent to the circle. Side BC is a secant intersecting the circle at point X.
What is the measure of angle ACB?

Answer 1

k

Answer 2

any suggestions?

Answer 3

king/gane?

Answer 4

Chumku?

Answer 5

First, draw a line from X to A. Notice that this creates an angle BXA that is also an inscribed angle. The degree of the arc it contains is 176. What is the measure of angle BXA?

Answer 6

half of that... 88 degrees?

Answer 7

the angle is inscribed, right?

Answer 8

Right. Now let me figure out where I was going with this again :(

Answer 9

you will get it :)

Answer 10

By the way, BAC is definitely an obtuse angle

Answer 11

so the angle we are looking for could not be 60 degrees

Answer 12

32°

60°

28°

16°

Answer 13

and I answered 16 degrees, which is also wrong

Answer 14

so only 28 degrees and 32 degrees are left :)

Answer 15

Think I got it. Don't know why I needed angle BXA. Note that 360-176=184. This is the measure of arc BXA.

Answer 16

k

Answer 17

92 degrees...

Answer 18

Since AC is tangential to the circle, we can think of it as an angle similar to the angle BXA we were looking at previously. That means, angle BAC should be \[184/2=92^{\circ}\]

Answer 19

so the angle we are looking for is 32 degrees :D:D:D:D:D:D:D::D:D:D:D!11111111

Answer 20

However, we want ACB and not BAC, so we look at \[180-56-92=32\]

Answer 21

Right.

Answer 22

k

Answer 23

can we do one more of these?