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In the figure shown, point O is the center of semicircle and [#permalink]

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In the figure shown, point O is the center of 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 segment 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

In the figure shown, point O is the center of 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 segment 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

In the figure shown, point O is the center of 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 segment 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 think 2 alone is sufficient

Given data : AB = OC which implies = BO.

So In triangle ABO, AB = BO , isoceles triangle hence opposite angles BOA =angle BAO similarly triangle

We know BO=OC and hence triangle BOC is also isoceles. So if angle BCO = 40, then angle OBC =40 and hence angle ABO = 180-40 = 120

once angle ABO is known, we know angle BOA = BAO, hence 2 angle BAO+ angle ABO = 180 2 BAO = 60 and hence BAO is 30

Hi, guys. If I follow your explanation correctly, I think you're assuming that AC is a straight line and that ACO is a triangle.

If that is what you're doing, is there anything in the question setup that says we can assume that AC is a straight line?

Thanks.

Good general question. You should almost assume that DS diagrams are not drawn to scale. But there are things about the picture that you can take as fact:

(1) If two lines look like they meet, they do. (i.e. no microscopic gaps) (2) If a line looks straight, it is. (i.e. no trick angles of 179.99 degrees) _________________

Emily Sledge | Manhattan GMAT Instructor | St. Louis

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