In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 113° and ∠HLK = 96°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 96° (Corresponding angles)
(c)
∠QRK
= 96° - 33.5°
= 62.5° (Exterior angle of a triangle)
Answer(s): (a) 33.5°; (b) 96°; (c) 62.5°