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by vladmire » Mon Dec 15, 2008 3:21 pm
I'm missing some fundamentals here, angle prs is directly horizontally across from the 90 degree angle right and angle pqr is directly above it. Where is angle PRQ? I was under the impression that prq and pqr where the same angle obviously not so... What needs to be learned here
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Source: — Data Sufficiency |
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by muzali » Mon Dec 15, 2008 4:55 pm
From the figure:
pqr + qps =90 =>pqr+ (qpr+rps) =90 ....(a)
and, rps + prs =90...(b)

We need to find |prs-pqr|

stm 1: qpr=30
from (a), pqr + 30 + rps =90, or rps = 60-pqr
from (b), rps =90 -prs
So, 60-pqr=90 -prs, hence we know the value of |prs-pqr|. Sufficient

stm 2: pqr + (180-prs) = 150, again this is sufficient to calculate |prs-pqr|.

Hence answer is D
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Angles

by vladmire » Mon Dec 15, 2008 5:25 pm
Can anyone elaborate just alittle more. I still don't know pqr or qps
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Re: Angles

by muzali » Mon Dec 15, 2008 5:44 pm
vladmire wrote:Can anyone elaborate just alittle more. I still don't know pqr or qps
Sorry, I should have explicity stated that pqr refers to the angle that is formed by line segments PQ and QR. Similarly, qps is the angle formed by line segments QP and PS.

Hope this helps.
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by pbanavara » Mon Dec 15, 2008 6:37 pm
You are asked to find : PRS - PQR

In any triangle : Exterior angle = sum of interior opposite angles

So in PQR : PRS is the exterior angle.

From 1: PRS = PQR + QPR => PRS -PQR = PRQ = 30 Sufficient

From 2: PQR + PRQ = 150.
PRQ = 180- PRS
PQR + 180 - PRS = 150 => PRS - PQR = 30

Hope this helps

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