In the figure, ∠WBF is a right-angled isosceles triangle. WF // YE , ∠DAZ = 50°, ∠BDC = 42° and ∠ZCA = 55°. Find
- ∠WFB
- ∠AYE
- ∠CED
(a)
∠WFB
= (180° - 90°) ÷ 2
= 90° ÷ 2
= 45° (Isosceles triangle)
(b)
∠BDX = 45° (Corresponding angles, FW//DX)
∠ADY
= ∠BDC + ∠BDX
= 42° + 45°
= 87°
∠AYE
= 180° - ∠DAZ - ∠ADY
= 180° - 50° - 87°
= 43° (Angles sum of triangle)
(c)
∠CDE
= 180° - ∠ADY
= 180° - 87°
= 93°(Angles in a straight line)
∠ECD = ∠ACB = 55° (Vertically opposite angles)
∠CED
= 180° - ∠ECD - ∠CDE
= 180° - 55° - 93°
= 32° (Angles sum of triangle)
Answer(s): (a) 45°; (b) 43°; (c) 32°