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Secondary 1 | Maths
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I am stuck. I would be grateful if u could help me out with this.
Date Posted: 4 years ago
Views: 340
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°
So your 1½a - 45° = 45°
1½a = 90°
a = 60°
PQR = 90° - a = 90° - 60° = 30°
= 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°
So your 1½a - 45° = 45°
1½a = 90°
a = 60°
PQR = 90° - a = 90° - 60° = 30°
Thanks so much. I get it now
It's quite interesting to note that as you go from the left most triangle to the rightmost, the apex angle in each triangle increases by 30°
Yes that's true
Doesn't seem like a sec 1 question though. Which level is this ?
In fact this question came up in SMO junior round 1. That one was a little easier but this question is a little tricky
No this is a sec 1 question
I am from NUS High so the curriculum varies
I see. I guessed right that it was Olympiad-style. Usually sec 1 syllabus for O level pathway won't have such questions
Oh, I see. Thanks for your help!
Welcome
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Will this suffice? Or am I missing anything. Thankyou so much!
Date Posted:
4 years ago
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
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°
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
The 16a - 1260° gives you -300° instead of 30° for PQR
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