In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 121° and ∠HLK = 102°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 121°) ÷ 2
= 29.5° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 102° (Corresponding angles)
(c)
∠QRK
= 102° - 29.5°
= 72.5° (Exterior angle of a triangle)
Answer(s): (a) 29.5°; (b) 102°; (c) 72.5°