In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 115° and ∠HLK = 101°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 115°) ÷ 2
= 32.5° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 101° (Corresponding angles)
(c)
∠QRK
= 101° - 32.5°
= 68.5° (Exterior angle of a triangle)
Answer(s): (a) 32.5°; (b) 101°; (c) 68.5°