NPQR is a square. CBQ and PBR are straight lines. PN = PC and ∠BPC = 11°. Find
- ∠PNC
- ∠RQC
.
(a)
∠NPR = 45° (Right angle)
∠NPC
= ∠NPR - ∠BPC
= 45° - 11°
= 34°
∠PNC
= (180° - ∠NPC) ÷ 2
= (180° - 34°) ÷ 2
= 146° ÷ 2
= 73° (Isosceles triangle)
(b)
PN = PC = PQ
CPQ is an isosceles triangle.
∠PCQ = ∠PQC (Isosceles triangle)
∠PQC
= (180° - ∠RPQ - ∠BPC) ÷ 2
= (180° - 45° -11°) ÷ 2
= 124° ÷ 2
= 62°
∠RQC
= ∠PQR - ∠PQC
= 90° - 62°
= 28°
Answer(s): (a) 73°; (b) 28°