PSLE In the figure, BCE and GFE are straight lines and FC is parallel to ED. ∠FGB is a right angle, ∠GBC = 52°, ∠BCD = 115° and ∠EFC = 68°.
- Find ∠BCF.
- Find ∠CDE.
(a)
∠BEG
= 180° - 90° - 52°
= 38° (Angles sum of triangle)
∠ECF
= 180° - 68° - 38°
= 74° (Angles sum of triangle)
∠BCF
= 180° - 74°
= 115° (Angles on a straight line)
(b)
∠ECD
= 360° - 115° - 74° - 115°
= 56° (Angles at a point)
∠CDE
= 180° - 56° - 56°
= 68°
Answer(s): (a) 115°; (b) 68°