In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 105° and ∠HLK = 98°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 105°) ÷ 2
= 37.5° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 98° (Corresponding angles)
(c)
∠QRK
= 98° - 37.5°
= 60.5° (Exterior angle of a triangle)
Answer(s): (a) 37.5°; (b) 98°; (c) 60.5°