2019 3marks P6 Paper2 Raffles Girls SA2 P6 / Raffles Girls / 2019 / SA2 / Paper 2 – Q08 [3 marks] Question: In the figure, ABCD is a square and EFGC is a trapezium. CE = BE, ∠EFG = 110° and ∠DCG = 230°.(a) Find ∠FGC.(b) Find ∠CBE. Solution: Similar Posts P6 / CHIJ / 2020 / SA2 Prelims / Paper 2 – Q11 [3 marks] P6 / Tao Nan / 2021 / SA2 Prelims / Paper 1 – Full Paper Solution
In case this solution needs more explanation, (a) Angle EFG + Angle FGC = 180 deg (Angles between parallel lines in trapezium = 180 deg) Angle FGC = 180 deg – Angle EFG = 180 deg – 110 deg = 70 deg (b) Angle FEC + Angle GCE =180 deg (Angles between parallel lines in trapezium = 180 deg) Angle GCE = 180 deg – Angle FEC = 180 deg – 90 deg = 90 deg Angle ECD = 360 deg – 230 deg – Angle GCE = 130 deg – 90 deg = 40 deg (Angles around a point = 360 deg) Angle BCD = 90 deg (right angle in a rect) = Angle ECB + Angle ECD Angle ECB = 90 deg – Angle ECD = 90 deg – 40 deg = 50 deg Reply
In case this solution needs more explanation,
(a) Angle EFG + Angle FGC = 180 deg (Angles between parallel lines in trapezium = 180 deg)
Angle FGC = 180 deg – Angle EFG = 180 deg – 110 deg = 70 deg
(b) Angle FEC + Angle GCE =180 deg (Angles between parallel lines in trapezium = 180 deg)
Angle GCE = 180 deg – Angle FEC = 180 deg – 90 deg = 90 deg
Angle ECD = 360 deg – 230 deg – Angle GCE = 130 deg – 90 deg = 40 deg (Angles around a point = 360 deg)
Angle BCD = 90 deg (right angle in a rect) = Angle ECB + Angle ECD
Angle ECB = 90 deg – Angle ECD = 90 deg – 40 deg = 50 deg