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26 Dec 2013, 06:01
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (01:10) correct
18%
(01:31)
wrong
based on 169
sessions
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In the figure above, the vertices of triangle ABC lie on the circumference of circle O. What is the value of x?
(A) 35
(B) 45
(C) 55
(D) 65
(E) 70
I'm puzzled. If one solve for angle AOB, it would imply a different result. Why?
GC_Circles-e1517384956903.png [ 10.87 KiB | Viewed 2583 times ]
Attachment:
GC_Circles-e1517384956903.png [ 10.87 KiB | Viewed 2583 times ]
26 Dec 2013, 06:20
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nechets
All angels:
Notice that OB = OC = radius, thus triangle OBC is isosceles. Therefore angle OCB = angle OBC --> x + x + 70° = 180° --> x = 55°.
Answer: C.
Hope it's clear.
Attachment:
Imagem1.jpg [ 27.04 KiB | Viewed 4380 times ]
31 Jan 2018, 00:50
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nechets
Solution
- • Assuming the O is the center of the circle and AC is the diameter.
- o We can infer that the lines OB = OC = OA = radius of the circle.
- o Since, it is an isosceles triangle
o Angle OBC = Angle OCB = x
o And we know that angle BOC = 70
- o Sum of internal angles of a triangle = 180
o Thus, for triangle OBC, we can write
- 70 + x + x = 180
2x = 110
x = 55
Regards,
Saquib
Quant Expert
e-GMAT
