PSLE In the figure, BCE and GFE are straight lines and FC is parallel to ED. ∠FGB is a right angle, ∠GBC = 52°, ∠BCD = 122° and ∠EFC = 62°.
- Find ∠BCF.
- Find ∠CDE.
(a)
∠BEG
= 180° - 90° - 52°
= 38° (Angles sum of triangle)
∠ECF
= 180° - 62° - 38°
= 80° (Angles sum of triangle)
∠BCF
= 180° - 80°
= 122° (Angles on a straight line)
(b)
∠ECD
= 360° - 122° - 80° - 122°
= 36° (Angles at a point)
∠CDE
= 180° - 36° - 36°
= 108°
Answer(s): (a) 122°; (b) 108°