Triangles and Semicircles

Good to know. Thanks.

masuarezdl wrote:

In the figure shown, point O is the center of the semicircle and points B, C, and D lie on the semicircle. If the length of line segment AB is equal to the length of line sement OC, what is the degree measure of angle BAO?

(1) The degree measure of angle COD is 60?.
(2) The degree measure of angle BCO is 40?.

I certainly dont see a way to solve this... Anybody?

What I am drawing from the facts and figure minus any statement is that if AB = OC then it's in fact AB = OC = BO = OD (each equal to radius); and if ∠BAO = x, then


∠ABO = 180 – 2 x

∠CBO = ∠BCO = 2 x

∠BOC = 180 – 4 x

∠COD = 180 – (x + 180 – 4 x) = 3 x.

(1) ∠COD = 60, or 3 x = 60, or x = 20. Sufficient

(2) ∠BCO = 40, or 2 x = 40, or x = 20. Sufficient

[spoiler]D[/spoiler]

mgshorrGMAT wrote:Someone please explain why CBO=2BAO...

I thought I was good at geometry until this problem.



Mark

An exterior angle of a triangle is equal to the sum of the opposite interior angles. So, if ABO is the triangle, then ∠CBO is one of its exterior angle, which is the sum of ∠BAO, and ∠BOA, we are free to treat ABC a straight line here. But ∠BAO = ∠BOA = x, hence ∠CBO = x + x = 2 x,

Or, ∠CBO = 2 ∠BAO.

But isn't there such a thing as: a triangle inscribed in a semicircle is always a right triangle?

If so, then <BOC is a right angle, then since triangle BOC is isosceles and right => <OBC=<OCB=45 degrees.


Based on this, the second option (<OCB=40 degrees) is still valid?

Thanks!

Brent@GMATPrepNow wrote:Here's my visual solution.
Please note that I only address why Statement (1) is sufficient, since there already seems to be agreement/understanding regarding the sufficiency of Statement (2)

Awesome Explanation Brent.

Your course and continuous support are spectacular.