In the figure AB and CD are two diameters of circle.
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- fskilnik@GMATH
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All angles are measured in degrees.
$$? = x = \angle CEB$$
01. Triangle AOC is isosceles with base AC, hence the 66-degrees angle is justified.
02. Angle BAC is inscribed in the circle, hence the 132-degrees red arc (CEB) is justified.
03. Angle x (our FOCUS) is inscribed in the circle and corresponds to the blue arc CADB, hence:
$$? = {{360 - 132} \over 2} = 114\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( A \right)$$
This solution follows the notations and rationale taught in the GMATH method.
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Fabio.
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Let's use some useful circle properties
First let's add a blue line, to divide ∠CEB into 2 angles
CD is the DIAMETER of the circle
Since ∠CED is an inscribed angle containing (aka "holding") the diameter, we can conclude that ∠CED = 90°
Now recognize that the intersection of AB and CD creates two equal (vertically opposite) angles, we can conclude the angle opposite the 48° is also 48°
Now recognize that we have two angles containing (aka "holding") the arc BD
Property: If a CENTRAL angle and an INSCRIBED angle contain the same arc, then the CENTRAL angle is TWICE the INSCRIBED angle
This means that ∠DEB = 24°
At this point, we have:
So, ∠CEB = 90° + 24° = 114°
Answer: A
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An INSCRIBED ANGLE is formed by two chords.
A CENTRAL ANGLE is formed by two radii.
When an inscribed angle and a central angle intercept the same two points on a circle, the inscribed angle is 1/2 the central angle.
Here, inscribed angle CFB and central angle COB both intercept the circle at points C and B.
Since central angle COB = 132º, inscribed angle CFB = (1/2)(132) = 66º.
Rule:
When a quadrilateral is inscribed in a circle, opposite angles sum to 180º.
Since opposite angles inside inscribed quadrilateral CEBF must sum to 180º, angle CEB = 180-66 = 114.
The correct answer is A.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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