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16 Jan 2013, 03:50
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Dear Mike
May be this problem is not the "GMAT" type but the following solution is less time consuming-
We draw QO which is a common tangent to both the circles (as in attached figure). Now, AQ=QC & BQ= QC (tangents from an external point of a circle are equal.
Let angle CAB = x & CBA=y
therefore, angle ACQ = x & BCQ = y
now, ACB+CAB+CBA=180 this implies,
(x+y)+ x + y = 180
i.e., x+y = 90 or angle ACB= 90
hope you find this helpful.
- Pari
May be this problem is not the "GMAT" type but the following solution is less time consuming-
We draw QO which is a common tangent to both the circles (as in attached figure). Now, AQ=QC & BQ= QC (tangents from an external point of a circle are equal.
Let angle CAB = x & CBA=y
therefore, angle ACQ = x & BCQ = y
now, ACB+CAB+CBA=180 this implies,
(x+y)+ x + y = 180
i.e., x+y = 90 or angle ACB= 90
hope you find this helpful.
- Pari
Attachments
Solution fig..pdf [81.45 KiB]
Downloaded 89 times
16 Jan 2013, 23:12
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In the figure below, AB, BC, BD & AC are chords of a circle.
The angle bisectors of \(\angle\)CBD & BCA intersect at E. EF is perpendicular to EC. If \(\angle CAB = 20^{\circ}\) and \(\angle BEF = \angle ABD\), then find\(\angle ACD\).
a. \(30^{\circ}\)
b. \(35^{\circ}\)
c. \(40^{\circ}\)
d. \(25^{\circ}\)
e. None of these
Sourece: T.I.M.E. CAT'11 Series
The angle bisectors of \(\angle\)CBD & BCA intersect at E. EF is perpendicular to EC. If \(\angle CAB = 20^{\circ}\) and \(\angle BEF = \angle ABD\), then find\(\angle ACD\).
a. \(30^{\circ}\)
b. \(35^{\circ}\)
c. \(40^{\circ}\)
d. \(25^{\circ}\)
e. None of these
Sourece: T.I.M.E. CAT'11 Series
Attachments
untitled.JPG [ 27.48 KiB | Viewed 4479 times ]
17 Jan 2013, 05:37
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I'm not explaining how I wrote the angles on the diagram. If you have doubts with that let me know. (use the property that the same arc will inscribe equal angles on the circle)
for ABCD - sum of angles is 360. so 4x+4y+2z+40=360. implies 4(x+y) +2z+40 = 360
for BEC sum of angles is 180. so y+z+x+90=180. implies x+y= 90-z ---> substitute this back up and you get z=20.
so E.
makes sense?
for ABCD - sum of angles is 360. so 4x+4y+2z+40=360. implies 4(x+y) +2z+40 = 360
for BEC sum of angles is 180. so y+z+x+90=180. implies x+y= 90-z ---> substitute this back up and you get z=20.
so E.
makes sense?
Attachments
Picture1.jpg [ 39.55 KiB | Viewed 4332 times ]
