NPQR is a square. DCQ and PCR are straight lines. PN = PD and ∠CPD = 15°. Find
- ∠PND
- ∠RQD
.
(a)
∠NPR = 45° (Right angle)
∠NPD
= ∠NPR - ∠CPD
= 45° - 15°
= 30°
∠PND
= (180° - ∠NPD) ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(b)
PN = PD = PQ
DPQ is an isosceles triangle.
∠PDQ = ∠PQD (Isosceles triangle)
∠PQD
= (180° - ∠RPQ - ∠CPD) ÷ 2
= (180° - 45° -15°) ÷ 2
= 120° ÷ 2
= 60°
∠RQD
= ∠PQR - ∠PQD
= 90° - 60°
= 30°
Answer(s): (a) 75°; (b) 30°