I think it's fine, though it's very long for a 4m question.

Edit : I made an error. It should be SRT = STR = (180° - 90°)/2 = 45 instead of SRT = TRS

This is my version of the answer working (with no assignment of letters to any angle)


QPR + TVU = 180° - PQV (angle sum of triangle PQV)

= 180° - 90° = 90°


Since QPR = QRP and TVU = VTU (base angles of isosceles triangle)

QRP + VTU also = 90°

PQR + TUV
= sum of angles in triangle PQR + sum of angles in triangle TUV - QPR - TVU - QRP - VTU

= 180° + 180° - 90° - 90°
= 180°

Since TUS + TUV = 180° (angles on a straight line),

This means TUS = PQR.

TSU = TUS (base angles of an isosceles triangle) = PQR


So TUS = TSU = PQR

Since RQS = RSQ, (base angles again)
and PQR + RQS = 90°(complementary angles),

Then TSU + RSQ = 90°



So RST = 180° - TSU - RSQ (angles on a straight line)
= 180° - 90°
= 90

So SRT = STR = (180° - RST)/2 (base angles of isosceles triangle)

= (180° - 90°)/2
= 45°


STR = TSU + TVU (exterior angle = sum of two interior opposite angles, wrt triangle TSV)


Now, TSU = TUS = TVU + VTU (exterior angle = sum of two interior opposite angles)
= 2TVU (since TVU = VTU)

So STR = 2TVU + TVU = 3 TVU

3TVU = 45°
TVU = 45° ÷ 3 = 15°

Since PQR = TUS = TSU = 2VTU ,

PQR = 2(15°) = 30°

I have already edited that while i was checking my work

Thanks for helping me

Thanks for Method 2.

Actually the sad part is that we are need to write a long winded explanation for even 1 mark question

A friend of mine is an alumnus of NUS High, he had similar experiences. Very rigorous curriculum

Hey wait. I spotted a 1440° - 16a in your working. Seems to not match the angle TUV. TUV is 150°, but 1440° - 16a gives you 480°

The 16a - 1260° gives you -300° instead of 30° for PQR