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 Post subject: GMAT GeometryPosted: Tue Oct 16, 2012 4:38 pm

Joined: Mon Nov 26, 2012 4:59 pm
Posts: 8

A circle is inscribed inside a right triangle as shown. If angle CAB = 60°, what is angle COE? (Note: Figure not drawn to scale.)
A. 120°
B. 135°
C. 150°
D. 165°
E. 180°

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

Therefore, angle COE = 360° – angle DAE – angle ACO – angle AEO.

We know that angle AEO is equal to 90° since the angle created by a radius and a tangent line is always 90°.
We also know that angle ACO is 45° because CDOF is a square and CO is its diagonal (CDOF is a square because all of its angles are 90° and DO = OF as radii).

Now we can solve for angle EOD:
Angle EOD = 360° – 60° – 45° – 90° = 165°.

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Why wouldn't angle ADO be 90 degrees (line DO is the radius and it hits a tangential line of the triangle)?

If it were 90 degrees then the answer would be 360 – 60 – 90 – 90 = 120 degrees

Thanks

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 Post subject: Re: GMAT GeometryPosted: Tue Oct 16, 2012 4:50 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
Why wouldn't angle ADO be 90 degrees (line DO is the radius and it hits a tangential line of the triangle)?
The angle ADO is 90 degrees indeed.

Quote:
If it were 90 degrees then the answer would be 360 – 60 – 90 – 90 = 120 degrees
Apparently, you have used angles in the quadrilateral ADOE and found the measure of the angle DOE. But the question asks for the angle COE, which equals angle DOE + angle DOC .

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