In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 117° and ∠HLK = 98°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 117°) ÷ 2
= 31.5° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 98° (Corresponding angles)
(c)
∠QRK
= 98° - 31.5°
= 66.5° (Exterior angle of a triangle)
Answer(s): (a) 31.5°; (b) 98°; (c) 66.5°