NPQR is a square. CBQ and PBR are straight lines. PN = PC and ∠BPC = 13°. Find
- ∠PNC
- ∠RQC
.
(a)
∠NPR = 45° (Right angle)
∠NPC
= ∠NPR - ∠BPC
= 45° - 13°
= 32°
∠PNC
= (180° - ∠NPC) ÷ 2
= (180° - 32°) ÷ 2
= 148° ÷ 2
= 74° (Isosceles triangle)
(b)
PN = PC = PQ
CPQ is an isosceles triangle.
∠PCQ = ∠PQC (Isosceles triangle)
∠PQC
= (180° - ∠RPQ - ∠BPC) ÷ 2
= (180° - 45° -13°) ÷ 2
= 122° ÷ 2
= 61°
∠RQC
= ∠PQR - ∠PQC
= 90° - 61°
= 29°
Answer(s): (a) 74°; (b) 29°
