ABCD is a square. XWC and BWD are straight lines. BA = BX and ∠WBX = 15°. Find
- ∠BAX
- ∠DCX
.
(a)
∠ABD = 45° (Right angle)
∠ABX
= ∠ABD - ∠WBX
= 45° - 15°
= 30°
∠BAX
= (180° - ∠ABX) ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(b)
BA = BX = BC
XBC is an isosceles triangle.
∠BXC = ∠BCX (Isosceles triangle)
∠BCX
= (180° - ∠DBC - ∠WBX) ÷ 2
= (180° - 45° -15°) ÷ 2
= 120° ÷ 2
= 60°
∠DCX
= ∠BCD - ∠BCX
= 90° - 60°
= 30°
Answer(s): (a) 75°; (b) 30°
