CDEF is a square. UTE and DTF are straight lines. DC = DU and ∠TDU = 15°. Find
- ∠DCU
- ∠FEU
.
(a)
∠CDF = 45° (Right angle)
∠CDU
= ∠CDF - ∠TDU
= 45° - 15°
= 30°
∠DCU
= (180° - ∠CDU) ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(b)
DC = DU = DE
UDE is an isosceles triangle.
∠DUE = ∠DEU (Isosceles triangle)
∠DEU
= (180° - ∠FDE - ∠TDU) ÷ 2
= (180° - 45° -15°) ÷ 2
= 120° ÷ 2
= 60°
∠FEU
= ∠DEF - ∠DEU
= 90° - 60°
= 30°
Answer(s): (a) 75°; (b) 30°
