# Triangle inscribed in circle

• Oct 10th 2010, 08:52 AM
BrownianMan
Triangle inscribed in circle
ABC inscribed within a circle whose diameter AC forms one of the sides of hte triangle. If Arc BC on the circle subtends an angle of 40 ddegrees, find the measure of angle BCA within the triangle.
• Oct 10th 2010, 09:01 AM
Plato
$m\angle ABC = 90^o$
• Oct 10th 2010, 09:15 AM
BrownianMan
So ABC is 90 degrees? Then if BAC is 40, BCA would have to be 50? But the answer says BCA is 70.
• Oct 10th 2010, 09:19 AM
Plato
How is an inscribed angle measured?
• Oct 10th 2010, 09:23 AM
BrownianMan
I'm not sure. I haven't done this since high school. Can you explain how to do so?
• Oct 10th 2010, 09:30 AM
Plato
$m\angle BAC = 20^o$
• Oct 10th 2010, 09:31 AM
BrownianMan
Isn't it 40? Why is it 20?
• Oct 10th 2010, 09:03 PM
sa-ri-ga-ma
If O is the center of the circle, then

$\angle(BOC) = 40^o$

$\angle(BOA) = 180^o - 40^o = 140^o = \angle(OBC) + \angle(OCB)$

But $\angle(OBC) = \angle(OCB)$

Now find CBA
• Oct 11th 2010, 03:21 AM
HallsofIvy
You've been given the measure of every angle in the triangle except the one you want!

An angle with its vertex on a circle subending an arc of $\theta$ degrees has measure $\frac{\theta}{2}$ degrees. Since BC subtends an arc of 40 degrees, angle BAC is 20 degrees. But you asked about angle BCA. There are two ways to get that.

1) The angles in a triangle sum to 180 degrees. Since AC is a diameter, angle ABC measures 90 degrees. We know that angle BAC is 20 degrees: 180- 90- 20= 180- 110= 70 degrees.

2) Arc ABC half the circle, 180 degrees. BC measures 40 degrees so arc AB measures 180- 40= 140 degrees. Angle ACB, then, is 140/2= 70 degrees.