The figure is not drawn to scale. HLNP is a parallelogram, GHP and HKP are isosceles triangles. PN//RX and ∠NVX= 101°. GL is straight line.
- Find ∠GPH.
- Find ∠LKP.
(a)
∠NVX = 101°
∠PTU = 101° (Corresponding angles)
∠STP
= 180° - 101°
= 79° (Angles on a straight line)
∠GPH
= 180° - 79° - 79°
= 22° (Isosceles triangle)
(b)
∠STP = ∠GHP = 79° (Corresponding angles)
∠PHK
= 180° - 79°
= 101° (Angles on a straight line)
∠HKU
= (180° - 101°) ÷ 2
= 39.5° (Isosceles triangle)
∠LKP
= 180° - 39.5°
= 140.5° (Angles on a straight line)
Answer(s): (a) 22°; (b) 140.5°