PSLE In the figure, BCE and GFE are straight lines and FC is parallel to ED. ∠FGB is a right angle, ∠GBC = 51°, ∠BCD = 114° and ∠EFC = 70°.
- Find ∠BCF.
- Find ∠CDE.
(a)
∠BEG
= 180° - 90° - 51°
= 39° (Angles sum of triangle)
∠ECF
= 180° - 70° - 39°
= 71° (Angles sum of triangle)
∠BCF
= 180° - 71°
= 114° (Angles on a straight line)
(b)
∠ECD
= 360° - 114° - 71° - 114°
= 61° (Angles at a point)
∠CDE
= 180° - 61° - 61°
= 58°
Answer(s): (a) 114°; (b) 58°