Pls help on GMAT Prep Q

Forgot to add the attachement

here it goes..

Angle R=90-T
Angle T=90-R

In triangle RQS

Since triangle RQS is isosceles angle Q= angle S (say A)
R+Q+S=180
90-T+2A=180
A=(90+T)/2


In triangle TSU

Since triangle TSU is isosceles angle U= angle S (say B)
T+S+U=180
90-R+2B=180
B=(90+R)/2


Now for line RST

angle RSQ + angle QSU + angle TSU = 180

A+X+B=180

(90+T)/2 + X + (90+R)/2 = 180

(90+T+2X+90+R)/2 = 180

(180+90+2X)/2 = 180 (Since angle R + angle T + 90 = 180 .....the triangle RPT)

(270+2X)/2 = 180

135+X = 180

X = 45

Hence the answer is C


Hope you got it!!!!

gautamberry wrote:here it goes..

Angle R=90-T
Angle T=90-R

In triangle RQS

Since triangle RQS is isosceles angle Q= angle S (say A)
R+Q+S=180
90-T+2A=180
A=(90+T)/2


In triangle TSU

Since triangle TSU is isosceles angle U= angle S (say B)
T+S+U=180
90-R+2B=180
B=(90+R)/2


Now for line RST

angle RSQ + angle QSU + angle TSU = 180

A+X+B=180

(90+T)/2 + X + (90+R)/2 = 180

(90+T+2X+90+R)/2 = 180

(180+90+2X)/2 = 180 (Since angle R + angle T + 90 = 180 .....the triangle RPT)

(270+2X)/2 = 180

135+X = 180

X = 45
Hence the answer is C


Hope you got it!!!!

or

B+x = A + R R=90-T
=> B + x = A + 90-T
and A + x = B + T

adding both we have
A+B+2x = A+B+90
therefore, 2x=90 or x=45

arvindm07 wrote:

gautamberry wrote:here it goes..

Angle R=90-T
Angle T=90-R

In triangle RQS

Since triangle RQS is isosceles angle Q= angle S (say A)
R+Q+S=180
90-T+2A=180
A=(90+T)/2


In triangle TSU

Since triangle TSU is isosceles angle U= angle S (say B)
T+S+U=180
90-R+2B=180
B=(90+R)/2


Now for line RST

angle RSQ + angle QSU + angle TSU = 180

A+X+B=180

(90+T)/2 + X + (90+R)/2 = 180

(90+T+2X+90+R)/2 = 180

(180+90+2X)/2 = 180 (Since angle R + angle T + 90 = 180 .....the triangle RPT)

(270+2X)/2 = 180

135+X = 180

X = 45
Hence the answer is C


Hope you got it!!!!

or

B+x = A + R R=90-T
=> B + x = A + 90-T
and A + x = B + T

adding both we have
A+B+2x = A+B+90
therefore, 2x=90 or x=45

What's the rule behind
B+x = A + R

LSB wrote:

arvindm07 wrote:

gautamberry wrote:here it goes..

Angle R=90-T
Angle T=90-R

In triangle RQS

Since triangle RQS is isosceles angle Q= angle S (say A)
R+Q+S=180
90-T+2A=180
A=(90+T)/2


In triangle TSU

Since triangle TSU is isosceles angle U= angle S (say B)
T+S+U=180
90-R+2B=180
B=(90+R)/2


Now for line RST

angle RSQ + angle QSU + angle TSU = 180

A+X+B=180

(90+T)/2 + X + (90+R)/2 = 180

(90+T+2X+90+R)/2 = 180

(180+90+2X)/2 = 180 (Since angle R + angle T + 90 = 180 .....the triangle RPT)

(270+2X)/2 = 180

135+X = 180

X = 45
Hence the answer is C


Hope you got it!!!!

or

B+x = A + R R=90-T
=> B + x = A + 90-T
and A + x = B + T

adding both we have
A+B+2x = A+B+90
therefore, 2x=90 or x=45

What's the rule behind
B+x = A + R

Exterior angle of a triangle is equal to the sum of the opposite interior angles.

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

While the angle is 90-T and not 180-90-T??!