More circle help!?

Question

The following is given:
*Circle O
*diameter AOB
*acute triangle ABC is inscribed in circle O, with the bottom of triangle ABC being diameter AOB.
*arc BC is equal to 80 degrees
Find the measure of angles ABC and ACB.Work please!Thank you!

**ABC is not necessarily an acute triangle. Sorry!**

Answer 1

The measure of Angle ACB must be 90 degrees because any triangle inscribed in a semicircle must be a right triangle. The measure of Angle ABC must be 50 degrees because the measure of Angle CAB is 40 degrees, or half of arc CB as an inscribed Angle. The angles of a triangle add up to 180 degrees, so 180 - 90 - 40 = 50 degrees.

Answer 2

The angle ACB is 90 degrees because it is inscribed in a semicircle. It equals1/2(180) = 90 degrees.

The angle ABc = 1/2 arc AC
Arc AC = 180 - arc BC = 180 - 80 = 100 degrees
So angle ABC = 1/2(100) = 50 degrees

Answer 3

An inscribed angle has half the measure of the arc it intercepts. Inscribed angle ACB is 90 degress since it intecepts a semicircle, or half the circle, which measures 180 degrees.

Inscribed angle CAB is 40 degrees since it intercepts arc BC which measures 80 degrees.

Triangle ABC has a 90 degree angle (right angle ACB) and a 40 degree angle (angle CAB). THe sum of the interior angles of a triangle is 180. Angle ABC is 50 degrees since 180 - 90 - 40 = 50.

So, angle ACB is 90 degrees and angle ABC is 50 degrees.

Answer 4

im not completely sure, its been a while since i worked with arc angles, but i think angle ABC is 50 degrees and angle acb is 90 degrees. Work is: draw the triangle in the circle, then you know that arc BC is 80 degrees, arc AC is 100 degrees, and the other half of the circle is 180 degrees. halve all of those measures to get the measures of the angles.

Answer 5

ABC = 50
ACB = 90
BAC = 40

First find Triangle BOC.
Angle BOC is 80 (from example)
Sides OB = OC (any point on the circle is defined as the radius from the origin)
Therefore, angle OBC and OCB are equivalent (by definition of Isosceles Triangle)
Since the summation of interior angles of a triangle is 180,
angle OBC + OCB + BOC = 180
OBC = OCB = 1/2 (180-BOC) = 50

Next find Triangle OAC:
Angle AOC is the supplementary angle of BOC = 180- BOC = 100 degrees
following the same concepts as above,
OAC = OCA = 1/2(180-100) = 40

Now apply knowledge to Angle ACB,
ACB = 180 - OAC - OBC = 180-50-40 = 90