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Triangle within a triangle and angles

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Triangle within a triangle and angles [#permalink] New post 06 Oct 2009, 11:15
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Hello, I am trying to understand the answer on this question. Any help would be great!!
Added a file with the question and drawing.
Hope you can see it. It is apparently there.
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Attachment:
TriangleQPS.JPG
TriangleQPS.JPG [ 2.5 KiB | Viewed 2079 times ]


In the figure shown, the measure of angle PRS is how many degrees greater than the measure of angle PQR.

(1) The measure of angle QPR is 30°

(2) The sum of the measures of angles PQR and PRQ is 150°
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Re: Triangle within a triangle and angles [#permalink] New post 06 Oct 2009, 11:51
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flehnen wrote:
Hello, I am trying to understand the answer on this question. Any help would be great!!
Added a file with the question and drawing.
Attachment:
triangle.pdf

Hope you can see it. It is apparently there.
thanks,


Q: <PRS-<PQR=?

(1) <QPR=30 --> in triangle QPR three angles sum=180=<QPR+<PQR+<PRQ -->180=30+<PQR+(180-<PRS) --> 30=<PRS-<PQR SUFFICIENT

(2) Basically the same information is given <PQR+<PRQ=150 --> <PQR+180-<PRS=150 --> 30=<PRS-<PQR SUFFICIENT

D.

OR:

(1) <QPR=30 --> an exterior angle of a triangle is equal to the sum of the opposite interior angles --> <PRS =<QPR + <PQR --> <PRS - <PQR = <QPR = 30. Sufficient.

Hope it helps.
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Re: Triangle within a triangle and angles [#permalink] New post 06 Oct 2009, 12:14
Got you! makes it obvious now.
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Re: Triangle within a triangle and angles [#permalink] New post 07 Oct 2009, 12:40
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Another solution -

Let LPRS = x and LPQR = y. Question is asking to find what is the difference between LPRS and LPQR.

IN triable PQR, LQPR and LPQR are the interior angles, LPRS is the exterior angle opposite to LQPR and LPQR. There is a rule that Exterior angle of a triangle = sum of the opposite interiod angles.

==> LPRS = LPQR + LQPR. ==> x = y + LQPR ==> x - y = LQPR. So if we know the angle of LQPR, we can solve the question.

Clue 1 = Given LQPR = 30. This is eactly what we need to solve the question. So sufficient.

Clue 2 = LPQR + LPRQ = 150 ==> From this we can deduce that LQPR= 30(since sum of all the interior angles is 180. Again we know LQPR and hence is sufficient to solve the question.

So Ans D.
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Re: Geometry [#permalink] New post 14 May 2010, 09:58
vannbj wrote:
I see how you can use both to get the answer but I don't get how you can use either one. Can someone help me out?


I dislike using the three letters to represent an angle. I prefer single variables.
Facts:
PRS = x
PQR = y
PRQ = z

z + x = 180
x + y + z = 180

We want to know the value of x-y

1. QPR = 30 degrees
If QPR = 30 then y + z = 150 (180-30)

y + z = 150
z + x = 180

z = 150 -y
substitute
150-y+x = 180
-y + x = 30
x - y = 30 so sufficient

2. statement two is saying y + z = 150 and that's the same equation we calculated on step 2 so sufficient
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Re: Triangle within a triangle and angles [#permalink] New post 15 May 2010, 08:21
great explanation Bunuel ....... Kudos
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Angles [#permalink] New post 18 Sep 2010, 12:38
Cn some one explain how the answer is D?
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Re: Angles [#permalink] New post 18 Sep 2010, 15:05
vigneshpandi wrote:
Cn some one explain how the answer is D?
Attachment:
Angles.JPG


Note that PRS+SPR=PQS+SPQ
So PRS-PQS=SPQ-SPR=QPR

1) Clearly sufficient, since it tells us what QPR is

2) QPR = 180 - (PQR + PRQ), so again we know QPR. Sufficient to find out out PRS-PQS

Hence answer is (d), either statement alone is sufficient
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 [#permalink] New post 18 Sep 2010, 15:11
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Known angle QSP = 90 degree. We need to find difference between angle PQR and PRS. Let PQR be x degree and PRS be y degree. We need the value of (y-x) degree.

Statement A: angle QPR is 30 degree.

Using the exterior angle concept, we know that angle QPR + angle PQR = angle PRS.

=> 30 + angle PQR = angle PRS => 30 + x = y => y-x = 30 degree. Hence we have got the required angle. Sufficient.

Statement B: angle PQR and PRQ = 150 degree.

In the triangle QPR, sum of the interior angles is 180 degree. That is angle QPR + angle PQR + angle PRQ = 180 degree.

Hence angle QPR = 180 - 150 = 30 degree. Here we have got the same angle as the one mentioned in statement A. Also we have proved that Statement A is sufficient. Hence Statement B is also sufficient.

Answer is D.
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Re: Angles [#permalink] New post 18 Sep 2010, 15:18
Thanks for the explanation...I am feeling very bad when I am unable to recollect the basic principles:(
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Re: Angles [#permalink] New post 18 Sep 2010, 15:24
vigneshpandi wrote:
Thanks for the explanation...I am feeling very bad when I am unable to recollect the basic principles:(
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Nothing to worry. With practice you should be able to recollect it comfortably. The trick is to not to look for the answer quickly but to keep prodding with the information presented in the question.
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Re: Triangle within a triangle and angles [#permalink] New post 16 Oct 2010, 09:36
grafical solution is always easier.
Answer is D.
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Re: Triangle within a triangle and angles   [#permalink] 16 Oct 2010, 09:36
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