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 Post subject: GMAT Geometry
PostPosted: Wed Jan 26, 2011 5:57 pm 
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Joined: Sun May 30, 2010 3:15 am
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In the figure above, a circle is inscribed in a right triangle. If angle CAB = 60°, what is the degree measure of angle EOD? (Note: Figure not drawn to scale.)
A. 120°
B. 135°
C. 140°
D. 150°
E. 155°

(A) Figure AEOD is a quadrilateral, so the sum of its interior angles must be 360°.

Therefore, angle EOD = 360° – angle DAE – angle ADO – angle AEO.

We know that both angles ADO and AEO are equal to 90° since the angle created by a radius and a tangent line is always 90°.

Now we can solve for angle EOD:
Angle EOD = 360° – 60° – 90° – 90° = 120°.

The correct answer is choice (A).
----------
Better Explanation?

Inscribed angle is half the arc it intercepts. Since both lines are tangent, angle CAB is an inscribed angle intersecting arc EOD, so by definition Angle EOD = 2 × 60 = 120


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 Post subject: Re: GMAT Geometry
PostPosted: Wed Jan 26, 2011 6:04 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
Unfortunately, the newly proposed solution is NOT correct.

The angle CAB is not an inscribed angle. An inscribed angle is the angle that is made by two chords (or a chord and a tangent in the confluent case). In our case the angle CAB is made by two tangents.

And the relation between the angle CAB and the angle DOE is:
angle DOE + angle CAB = 180°
So if angle CAB was 50°, then the angle DOE would be 130° (NOT 100°).


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 Post subject: Re: GMAT Geometry
PostPosted: Tue Nov 15, 2011 11:32 am 
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When you state that "the relation between the angle CAB and the angle DOE is:angle DOE + angle CAB = 180°" please confirm that you know this because angles ADO and OEA each are 90° (which we know because ADC and AEB are each perpendicular to the radii bc inscribed) leaving 180 remaining degrees of the quadrilateral. If there is another principle behind this relationship, please explain. Also, is there any other way to solve this problem without identifying the quadrilateral?

Thanks.


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 Post subject: Re: GMAT Geometry
PostPosted: Wed Nov 16, 2011 11:40 am 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
Student wrote:
When you state that "the relation between the angle CAB and the angle DOE is:angle DOE + angle CAB = 180°" please confirm that you know this because angles ADO and OEA each are 90° (which we know because ADC and AEB are each perpendicular to the radii bc inscribed) leaving 180 remaining degrees of the quadrilateral.
That is correct.

Quote:
Also, is there any other way to solve this problem without identifying the quadrilateral?
I see no clearer way to solve this question. Basically, we've solved it in two steps:
1. The sum of all angles in the quadrilateral is 360°.
2. One of the angles is given, 60°. Another two are 90°, because of the tangent lines. So the remaining angle is 360° - 90° - 90° - 60° = 120°.

If you didn't know the property of angles in a quadrilateral (that they sum up to 360⁰), then you could divide the quadrilateral into two triangles. It can be divided either into two equal right triangles, or into two isosceles triangles, depending on which diagonal you draw. In each case you would also need to use the fact that tangent lines make 90⁰ angles with the radii.


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