In the figure, HJKL and JNPQ are two rhombuses. If ∠NPQ = 112° and ∠HLK = 97°, calculate
- ∠PNQ
- ∠PKJ
- ∠QRK
(a)
∠PNQ
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
(b)
∠PKJ = ∠HLK = 97° (Corresponding angles)
(c)
∠QRK
= 97° - 34°
= 63° (Exterior angle of a triangle)
Answer(s): (a) 34°; (b) 97°; (c) 63°