In the figure, HJKL is a trapezium and HLMQ is a rhombus. HL is parallel to JK and LM = LN. QLK and LPN are straight lines. ∠LHQ = 52° and ∠PLQ = 28°. Find
- ∠HLK
- ∠LNM
(a)
∠HLQ
= (180° - 52°) ÷ 2
= 64° (Isosceles triangle)
∠HLK
= 180° - 64°
= 116° (Angles on a straight line)
(b)
∠NLM
= 64° - 28°
= 36°
∠LNM
= (180° - 36°) ÷ 2
= 72 ° (Isosceles triangle)
Answer(s): (a) 116°; (b) 72°