Triangles and Semicircles
- tienvunguyen
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- sanju09
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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, thenmasuarezdl 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?
∠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]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
- sanju09
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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,mgshorrGMAT wrote:Someone please explain why CBO=2BAO...
I thought I was good at geometry until this problem.
Mark
Or, ∠CBO = 2 ∠BAO.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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!
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!
- Ashishkapoor7
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