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85%
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35% (02:49) correct
65%
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based on 23
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An acute isosceles triangle, ABC is inscribed in a circle. Through B and C, tangents to the circle are drawn, meeting at point D. If \(\angle\)ABC = \(\angle\)ACB = 2* \(\angle\)D and x is the radian measure of \(\angle\)A, then x =
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JeffTargetTestPrep
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Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,973
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert reply
Posts: 2,973
twobagels
angle ACB = 2(angle D), so if angle D = y, then angle ABC = ACB = 2y.
If O denotes the center of the circle, then following must be true about the interior angles of deltoid OBDC:
angle COB + angle OBD + angle D + angle OCD = 360 [Because the sum of the interior angles of any quadrilateral is always equal to 360 degrees.]
angle COB + 90 + y + 90 = 360 [Two angles are 90 degrees because each of them is defined by a tangent and a radius.]
angle COB = 180 – y
Since angle A is an inscribed angle to minor arc CB and angle COB is a central angle to minor arc CB, we have:
angle A = angle COB/2
angle A = (180 – y)/2
For triangle ABC, we have:
(180 – y)/2 + 2y + 2y = 180
180 – y + 8y = 360
y = 180/7 degrees
Therefore,
angle A = (180 – 180/7)/2 degrees
Since 180 degrees is equal to pi radian, we have:
x = angle A = (pi – pi/7)/2 radian = 3pi/7 radian
Answer: A
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General Discussion
GMAT 1: 720 Q50 V36
Posts: 234
Bunuel does any other method exist to solve this apart from as shown by JeffTargetTestPrep in which it can be solved based on center approach, however a bit lengthy.
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