I think the answer is c.
let <QRS = t. and <STU = w.
Now we know that t+w = 90 ..... (1)
From stmt 1 :: RQS is isosceles triangle so <RQS = <RSQ = 90-(t/2)
From stmt 2 :: STU is isosceles riangle so < TSU = <TUS> 90-(w/2) + 90-(t/2) + x = 180 .... (2)
Lastly QSUP is a quadrilateral.
<PQS = 180 - 90 + (t/2)
<PUS = 180 - 90 + (w/2)
So 180 - 90 + (t/2) + 180 - 90 + (w/2) + 90 + x = 360 ..... (3).
So we have 3 unknowns and three equations.
Hence we can get x.
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The explanation doesn't help at all, Can some post another approach to solve this problem?
1 says- QR = RS
hence angle RQS = angle RSQ, let angle RQS = angle RSQ = y.
in triangle RQS,
y + y + angle QRS = 180 degree
-> angle QRS = 180 degree - 2y ---- eqn i
now, from 2 we get,
ST = TU
-> angle SUT = angle STU. let angle SUT = angle STU = z
in triangle STU
z + z + angle STU = 180 degree
-> angle STU = 180 - 2z --- eqn ii
now, in triangle RPT,
angle RTP + angle STU + angle PRS = 180 degree
-> 90 degree + (180 - 2z) + (180 degree - 2y) = 180 degree ( solved from eqn i & ii)
-> y+ z = 136 degree -- eqn iii
now,
angle x + angle y + angle z = 180 degree
-> angle x = 180 - 135 (solved from eqn iii)
therefore angle x = 45 degree.
hope it helps
hence angle RQS = angle RSQ, let angle RQS = angle RSQ = y.
in triangle RQS,
y + y + angle QRS = 180 degree
-> angle QRS = 180 degree - 2y ---- eqn i
now, from 2 we get,
ST = TU
-> angle SUT = angle STU. let angle SUT = angle STU = z
in triangle STU
z + z + angle STU = 180 degree
-> angle STU = 180 - 2z --- eqn ii
now, in triangle RPT,
angle RTP + angle STU + angle PRS = 180 degree
-> 90 degree + (180 - 2z) + (180 degree - 2y) = 180 degree ( solved from eqn i & ii)
-> y+ z = 136 degree -- eqn iii
now,
angle x + angle y + angle z = 180 degree
-> angle x = 180 - 135 (solved from eqn iii)
therefore angle x = 45 degree.
hope it helps
Last edited by rabab on Wed Sep 10, 2008 10:13 am, edited 1 time in total.
