Bunuel wrote:

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°

Kudos for a correct solution. We see that angle EOD is a central angle with intercepted arc DE. We know that the measure of a central angle is equal to its intercepted arc, and thus if we can determine the measure of arc DE, then we can determine the measure of angle EOD.

We are given that angle CAB = 60°. Since angle CAB is formed by two tangents to circle O (tangents AB and AC), the measure of angle CAB = ½ (arc DFE - arc DE). However, arc DFE + arc DE = 360°, and thus arc DFE = 360° - arc DE. We can substitute this back in the equation:

angle CAB = ½ (arc DFE - arc DE)

60° = ½ (360° - arc DE - arc DE)

60° = ½ (360° - 2(arc DE))

60° = 180° - arc DE

arc DE = 120°

Since arc DE = 120°, angle EOD = 120° also.

Alternate Solution:

Since the circle is tangent to the triangle at points D and E, the angles formed between the sides of the triangle and the radii of the circle at these points is 90°; in other words, angle ADO = angle AEO = 90°.

Let’s look at the quadrilateral ADOE. We know the sum of the interior angles is 360° and we know three of its angles are 60°, 90° and 90°. Thus, the remaining angle, which is the angle AOE is 360 - (90 + 90 + 60) = 120°.

Answer: A

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Jeffery Miller

Head of GMAT Instruction

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