BCDE is a square. DCD and CCE are straight lines. CB = CD and ∠CCD = 11°. Find
- ∠CBD
- ∠EDD
.
(a)
∠BCE = 45° (Right angle)
∠BCD
= ∠BCE - ∠CCD
= 45° - 11°
= 34°
∠CBD
= (180° - ∠BCD) ÷ 2
= (180° - 34°) ÷ 2
= 146° ÷ 2
= 73° (Isosceles triangle)
(b)
CB = CD = CD
DCD is an isosceles triangle.
∠CDD = ∠CDD (Isosceles triangle)
∠CDD
= (180° - ∠ECD - ∠CCD) ÷ 2
= (180° - 45° -11°) ÷ 2
= 124° ÷ 2
= 62°
∠EDD
= ∠CDE - ∠CDD
= 90° - 62°
= 28°
Answer(s): (a) 73°; (b) 28°
