In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 122° and ∠HLK = 92°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 122°) ÷ 2
= 29° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 92° (Corresponding angles)
(c)
∠QRK
= 92° - 29°
= 63° (Exterior angle of a triangle)
Answer(s): (a) 29°; (b) 92°; (c) 63°