BCDE is a square. XWD and CWE are straight lines. CB = CX and ∠WCX = 11°. Find
- ∠CBX
- ∠EDX
.
(a)
∠BCE = 45° (Right angle)
∠BCX
= ∠BCE - ∠WCX
= 45° - 11°
= 34°
∠CBX
= (180° - ∠BCX) ÷ 2
= (180° - 34°) ÷ 2
= 146° ÷ 2
= 73° (Isosceles triangle)
(b)
CB = CX = CD
XCD is an isosceles triangle.
∠CXD = ∠CDX (Isosceles triangle)
∠CDX
= (180° - ∠ECD - ∠WCX) ÷ 2
= (180° - 45° -11°) ÷ 2
= 124° ÷ 2
= 62°
∠EDX
= ∠CDE - ∠CDX
= 90° - 62°
= 28°
Answer(s): (a) 73°; (b) 28°
