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CATPREP 3 - geometry

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by gbb » Sat Mar 21, 2009 5:08 pm
answer d.
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by DanaJ » Sat Mar 21, 2009 11:20 pm
This problem requires some knowledge about inscribed angles in a circle.

1. notice that angle AOB and angle ACB subtend the same minor arc: AB. Since this is the case, then ACB's measurement is a piece of cake (provided you know the theorem, of course). ACB will be 2 times smaller than AOB, since ACB has its vertex on the circle and AOB has its vertex in the origin. This makes ACB 75 degrees.

2. Again, we use the property outlined above, except that in this case we need to find out the measurement of AOB.
AOB subtends minor arc AB of 10pi/3. The total length of the circle is 2Pi*radius = 2*4*Pi = 8Pi and corresponds to the whole circle of 360 degrees. You get:
8Pi...............360 degrees
10Pi/3...........x degrees

x will be [(10Pi/3)*360]/8Pi = 150 degrees. This is why AOB is again 150 degrees and ACB will be half that measurement.
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by gbb » Sun Mar 22, 2009 7:47 am
Thanks DanaJ.

Sometimes I get stuck with geometry. Do you know where can I find a good material explaining all the theory? (mainly circles, triangles and triangles+anyform).

Thanks again.
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by DanaJ » Sun Mar 22, 2009 8:26 am
Sorry, unfortunately I don't. I've only worked with OG so far in my prep, and their math section is not very detailed.
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by cubicle_bound_misfit » Sun Mar 22, 2009 8:31 am
Answer is D

Both statement lead to finding inscribed angle AOB.
Cubicle Bound Misfit
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