As QR=RS, <RQS=<RSQ. Similarly as TS=TU, <TSU=<TUS.----------- (1)

Also, <RSQ+<x+<TSU=180 degrees.
Using (1) <RQS +x+<SUT=180 degrees.--------------------(2)

Now sum of angles of a quadrilateral is 360 degrees.
Hence <x+<SQP+<SUP=270 degrees.---------------------------(3)

<RQS+<SQP=180 degrees----------------------------(4)
<SUT+<SUP=180 degrees----------------------------(5)


Using (2), (4) & (5)..... 180 - <SQP+ x + 180-<SUP = 180

180+X=<SQP+<SUP--------------------------(6)

Using (3) & (6)

180+x=270-x

x=45 degrees.......Hence answer is C.

Am not sure if there is any easy way to solve this question



Sukrant........

thanks!

I think this might be an easier method?

After redrawing the triangle to fit assumptions (1) and (2), I realized that the entire triangle must actually be an isosceles triangle (someone please correct me if I am wrong). Accordingly, angles QRS and UTS each equal 45 degrees. From there, angles QSR and UST are solved to equal 67.5 degrees each, leaving angle x to equal 180 - (67.5 + 67.5) = 45 degrees.

Think this might be a shorter way...

Let's call angle SUT and TSU "y"
Let's call angle RQS and RSQ "p"

Since angle SUT and TSU are equal, angle STU is 180-2y
Since RQS and RSQ are equal, angle QRS is 180 -2p

But we also know that angle QRS = 180-(RPT+PTR)
180-2p= 180-(90+180-2y)
2y +2p = 270
y+p=135

Therefore x= 180-135=45