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In the figure above, a circle is inscribed in a right triangle. If

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In the figure above, a circle is inscribed in a right triangle. If [#permalink]

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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°

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
geometry.gif
geometry.gif [ 3.03 KiB | Viewed 1416 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, a circle is inscribed in a right triangle. If [#permalink]

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New post 21 Jul 2015, 02:45
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Expert's post
Bunuel wrote:
Image
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.

[Reveal] Spoiler:
Attachment:
The attachment geometry.gif is no longer available


Check Figure

Answer: option A
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In the figure above, a circle is inscribed in a right triangle. If [#permalink]

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New post 21 Jul 2015, 03:24
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Bunuel wrote:
Image
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.

[Reveal] Spoiler:
Attachment:
geometry.gif


An "incircle" in a right angled triangle has the sides of the triangle tangents to the radii such that the radii are perpendicular to the tangents at the point of tangency. What this means is that the radii OE and OD are perpendicular to sides AB and AC respectively and thus \(\angle {ADO}\) and \(\angle {OEA}\) are both 90 degrees.

Thus in quadrilateral ADOE,

\(\angle{DAE} + \angle{ADO} + \angle{OEA} + \angle{DOE} = 360\)---> 60+90+90+x = 360 ---> x = 120 degrees. A is the correct answer.

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Re: In the figure above, a circle is inscribed in a right triangle. If [#permalink]

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New post 26 Jul 2015, 10:40
Bunuel wrote:
Image
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.

[Reveal] Spoiler:
Attachment:
geometry.gif


800score Official Solution:

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).
_________________

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the figure above, a circle is inscribed in a right triangle. If [#permalink]

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New post 13 Oct 2017, 08:24
Bunuel wrote:
Image
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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Re: In the figure above, a circle is inscribed in a right triangle. If   [#permalink] 13 Oct 2017, 08:24
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