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POWERPREP: Semicircle

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Senior Manager
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POWERPREP: Semicircle [#permalink] New post 09 Dec 2003, 13:58
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Senior Manager
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 [#permalink] New post 09 Dec 2003, 17:43
Some one needs to provide a basis for discussion so here goes ..

is it C?

Now why am I not confident here!!!
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 [#permalink] New post 09 Dec 2003, 17:44
D.
trigonometry:
Q is 90 degrees <angle made by diameter inside the semi-circle>
find different angles of the triangle.
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 [#permalink] New post 09 Dec 2003, 17:46
How can you find the other angles?
Please explain ...
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 [#permalink] New post 09 Dec 2003, 17:52
pitts20042006 wrote:
How can you find the other angles?
Please explain ...


from A
tanP = 1/2; you know P, then angle PQX = 90-P
thus, RQX = 90 - PQX; you will know R from here. Calculate XR,
tanR = QX/XR

same for B.
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 [#permalink] New post 09 Dec 2003, 17:56
oooo .. yaaaaaaa
I knew there was something that was not clicking with C .. spending too much time on verbal .. need to brush up with some quant as well.
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 [#permalink] New post 10 Dec 2003, 14:46
dj wrote:
pitts20042006 wrote:
How can you find the other angles?
Please explain ...


from A
tanP = 1/2; you know P, then angle PQX = 90-P
thus, RQX = 90 - PQX; you will know R from here. Calculate XR,
tanR = QX/XR

same for B.


Well done, answer is D. Actually I get tanP=a/2=4/2=2 and not 1/2, but the answer is the same.


I thought GMAT didn't test trigonometry. Does anybody know a different way of solving this?
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 [#permalink] New post 10 Dec 2003, 14:46
Aren't both triangles "special triangles" or something?
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 [#permalink] New post 10 Dec 2003, 14:48
stoolfi wrote:
Aren't both triangles "special triangles" or something?


well .. yeah.. they are right triangles... :wink:
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 [#permalink] New post 10 Dec 2003, 14:51
30-60-90 triangles!

hence, special...
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 [#permalink] New post 10 Dec 2003, 14:57
stoolfi wrote:
30-60-90 triangles!

hence, special...



Nah ah... if a triangle has 30-60-90 angles THEN it is a right triangle, it doesn't work the other way. Not all right triangles have 30-60-90 angles.
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 [#permalink] New post 10 Dec 2003, 15:11
Once again. Can we solve this question without using trignometry?
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 [#permalink] New post 10 Dec 2003, 16:54
preyshi wrote:
Once again. Can we solve this question without using trignometry?


i coudn't think of anything else. this is the easiest way, I presume.
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 [#permalink] New post 11 Dec 2003, 00:14
preyshi wrote:
Once again. Can we solve this question without using trignometry?


1) these are NOT 30-60-90 triangles,

2) you don't care. just use properties of SIMILAR triangles. There are 3 of them in the diagram.
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 [#permalink] New post 12 Dec 2003, 06:20
preyshi wrote:
Once again. Can we solve this question without using trignometry?

You can solve this using similiarity principles.

Let's call the point where segment a and b meet point Z.

(1) PZ = a = 4
so the similiarity
PZ/QZ = QZ/ZR
4/QZ = QZ/ZR
ZR=QZ┬▓/4 = 2┬▓/4
ZR = b = 1

(2) ZR = b = 1
so the similiarity
ZR/QZ=QZ/PZ
1/QZ = QZ/PZ
PZ = QZ┬▓ = 2┬▓
PZ = a = 4
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 [#permalink] New post 21 Dec 2003, 11:45
Just thought it might help to explain how we choose the simillar triangles. Taking notation from wonder_gmats solution

Triangle PQR is simillar to triangle QZR. We use Angle side angle as the simillarity law. Size QZ is common. 90 degrees is common to both. now the Angle ZPQ = Angle ZQR. So prove this you have to know that angle PQR = 90 degrees.
From this you can deduce that angle ZPQ = angle ZQR, hence we have angle side angle as your simillarity criteria.

I hope this helps. Anyone has shorter way to prove the same ?
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 [#permalink] New post 22 Dec 2003, 17:43
I think this is another way of solving this using Pythagore (it's also longer than the previous methods):

1) You have a=4

Thus PQ = v(4^2 + 2^2) = v20
(a + b)^2 = (v20)^2 + QR^2
But QR^2 is also equal to: (2^2 + b^2)
Substituting for QR we get: (4+b)^2 = (v20)^2 + (2^2 + b^2)

Now solving it gives us:
16+8b+b^2 = 20 + 4 + b^2
8b = 24 - 16 = 8
b = 1

The same logic would apply to statement 2 and the answer would be D, that you can solve it with either statement. Also, this is assuming that you know that a triangle inscribed within a semi-circle is a right triangle.
  [#permalink] 22 Dec 2003, 17:43
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