In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 113° and ∠HLK = 92°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 92° (Corresponding angles)
(c)
∠QRK
= 92° - 33.5°
= 58.5° (Exterior angle of a triangle)
Answer(s): (a) 33.5°; (b) 92°; (c) 58.5°