PSLE In the figure, BCE and GFE are straight lines and FC is parallel to ED. ∠FGB is a right angle, ∠GBC = 52°, ∠BCD = 112° and ∠EFC = 66°.
- Find ∠BCF.
- Find ∠CDE.
(a)
∠BEG
= 180° - 90° - 52°
= 38° (Angles sum of triangle)
∠ECF
= 180° - 66° - 38°
= 76° (Angles sum of triangle)
∠BCF
= 180° - 76°
= 112° (Angles on a straight line)
(b)
∠ECD
= 360° - 112° - 76° - 112°
= 60° (Angles at a point)
∠CDE
= 180° - 60° - 60°
= 60°
Answer(s): (a) 112°; (b) 60°