NPQR is a square. ZYQ and PYR are straight lines. PN = PZ and ∠YPZ = 13°. Find
- ∠PNZ
- ∠RQZ
.
(a)
∠NPR = 45° (Right angle)
∠NPZ
= ∠NPR - ∠YPZ
= 45° - 13°
= 32°
∠PNZ
= (180° - ∠NPZ) ÷ 2
= (180° - 32°) ÷ 2
= 148° ÷ 2
= 74° (Isosceles triangle)
(b)
PN = PZ = PQ
ZPQ is an isosceles triangle.
∠PZQ = ∠PQZ (Isosceles triangle)
∠PQZ
= (180° - ∠RPQ - ∠YPZ) ÷ 2
= (180° - 45° -13°) ÷ 2
= 122° ÷ 2
= 61°
∠RQZ
= ∠PQR - ∠PQZ
= 90° - 61°
= 29°
Answer(s): (a) 74°; (b) 29°
