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In the figure shown, point O is the center of the semicircle [#permalink]
15 Dec 2009, 12:09

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55% (02:26) correct
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In the figure shown, point O is the center of the semicircle and points B, C, 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 60º. (2) The degree measure of angle BCO is 40º.

In the figure shown, point O is the center of the semicircle [#permalink]
15 Dec 2009, 13:04

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In the figure shown, point O is the center of the semicircle and B, C, 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 ?

Write down everything you know from the stem:

\(BO=CO=radius=AB\) --> triangles BOC and ABO are isosceles. \(\angle BAO = \angle BOA\) and \(\angle BCO = \angle CBO\) \(\angle CBO = 2*\angle BAO\)

(1) The degree measure of angle COD is 60º: \(\angle BAO +\angle ACO = \angle COD = 60º\) degrees (Using exterior angle theorem) \(\angle ACO = \angle CBO = 2* \angle BAO\) \(So, \angle BAO + \angle ACO = 2* \angle BAO + \angle BOA = 3* \angle BAO = 60º\) \(\angle BAO = 20º\). SUFFICIENT

(2) The degree measure of angle BCD is 40º: \(\angle BCO=40º\) --> \(\angle BCO = \angle CBO=40º = 2*\angle BAO\) --> \(\angle BAO=20º\). SUFFICIENT

Re: In the figure shown, point O is the center of the semicircle [#permalink]
18 Dec 2009, 17:33

Given OC = AB OC = OB since both are radii. Therefore OB = AB. Since angles opposite to equal sides are equal. Therefore, Angle OCB = Angle OBC and Angle OAB = Angle BOA

Statement 1 Angle COD = 60 degrees. Not sufficient. Since we cannot determine angle COB or angle OCB. Statement 2 Angle OCB = 40 degrees. Not sufficient Since we cannot determine angle COB

Both together helps us determine Angle COB and angle BOA = 180 - angle COD - angle COB Therefore, answer is C _________________

Re: In the figure shown, point O is the center of the semicircle [#permalink]
23 Jul 2014, 01:13

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Re: In the figure shown, point O is the center of the semicircle [#permalink]
30 Jun 2015, 19:07

Bunuel wrote:

In the figure shown, point O is the center of the semicircle and B, C, 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 60º. (2) The degree measure of angle BCD is 40º.

Write down everything you know from the stem:

BO=CO=r=AB --> BOC and ABO are isosceles. <BAO=<BOA and <BCO=<OBC <CBO=2<BAO

Re: In the figure shown, point O is the center of the semicircle [#permalink]
01 Jul 2015, 00:24

Expert's post

honchos wrote:

Bunuel wrote:

In the figure shown, point O is the center of the semicircle and B, C, 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 ?

Write down everything you know from the stem:

\(BO=CO=radius=AB\) --> triangles BOC and ABO are isosceles. \(\angle BAO = \angle BOA\) and \(\angle BCO = \angle CBO\) \(\angle CBO = 2*\angle BAO\)

(1) The degree measure of angle COD is 60º: \(\angle BAO +\angle ACO = \angle COD = 60º\) degrees (Using exterior angle theorem) \(\angle ACO = \angle CBO = 2* \angle BAO\) \(So, \angle BAO + \angle ACO = 2* \angle BAO + \angle BOA = 3* \angle BAO = 60º\) \(\angle BAO = 20º\). SUFFICIENT

(2) The degree measure of angle BCD is 40º: \(\angle BCO=40º\) --> \(\angle BCO = \angle CBO=40º = 2*\angle BAO\) --> \(\angle BAO=20º\). SUFFICIENT

Answer: D. .

What property have you used Bunuel - <CBO=2<BAO

Triangle Exterior Angle Theorem: An exterior angle of a triangle is equal to the sum of the opposite interior angles. \(\angle CBO = \angle BAO + \angle BOA\) and since we know that \(\angle BAO = \angle BOA\), then \(\angle CBO = 2*\angle BAO\).

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