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°.

The correct answer is D.

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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