In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 107° and ∠HLK = 97°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 107°) ÷ 2
= 36.5° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 97° (Corresponding angles)
(c)
∠QRK
= 97° - 36.5°
= 60.5° (Exterior angle of a triangle)
Answer(s): (a) 36.5°; (b) 97°; (c) 60.5°