In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 123° and ∠HLK = 95°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 123°) ÷ 2
= 28.5° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 95° (Corresponding angles)
(c)
∠QRK
= 95° - 28.5°
= 66.5° (Exterior angle of a triangle)
Answer(s): (a) 28.5°; (b) 95°; (c) 66.5°