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GMAT Prep - Geometry

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by anshul265 » Tue Mar 04, 2008 1:52 pm
let angle qrs = s1
also qsr = s1

let angle tsu = s2
also sut = s2

in the rt angle triangle R + T + 90 = 180
therefore R + T = 90
now:

s1 + s2 + x = 180 (1)

2s1 + R = 180 (2)
2s2 + T = 180 (3)

2+3

2(s1+s2) +R + T = 360

therefore s1 + s2 = 135

substitute this in 1 ... calc x.

Hence answer C
-AN
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by abhi75 » Mon Mar 10, 2008 7:29 am
For 10 of 37,
Draw a line segment PR where R is on the X axis and PR is perpendicular to X axis.

Now, PRO is a right triangle with PR = 1 and OR = sqrt(3)
OP = sqrt(PR^2 + OR^2) = sqrt(1+3) = 2.
This is a 30-69-90 triangle (its a right triangle with ratios of sides as 1:sqrt3:2)

Angle POR = 30 (since it's opposite to side of length 1 in the above mentioned triangle).

Now, draw a line segment QT where T is on the X axis and QT is perpendicular to X axis.
POR + POQ + QOT = 180 (they form a straight line)
30 + 90 + QOT = 180
QOT = 60
In triangle OQT, QOT = 60, QTO = 90. So, OQT = 30

Also, OP = OQ = 2 (radius of the circle)
So, again from the 30-60-90 triangle property, since OQ = 2, OT = 1 and TQ = sqrt(3).

Hence s = 1.

So the answer is B.
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