PSLE In the figure, BCE and GFE are straight lines and FC is parallel to ED. ∠FGB is a right angle, ∠GBC = 53°, ∠BCD = 115° and ∠EFC = 64°.
- Find ∠BCF.
- Find ∠CDE.
(a)
∠BEG
= 180° - 90° - 53°
= 37° (Angles sum of triangle)
∠ECF
= 180° - 64° - 37°
= 79° (Angles sum of triangle)
∠BCF
= 180° - 79°
= 115° (Angles on a straight line)
(b)
∠ECD
= 360° - 115° - 79° - 115°
= 51° (Angles at a point)
∠CDE
= 180° - 51° - 51°
= 78°
Answer(s): (a) 115°; (b) 78°