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 = 48° and ∠PLQ = 31°. Find
- ∠HLK
- ∠LNM
(a)
∠HLQ
= (180° - 48°) ÷ 2
= 66° (Isosceles triangle)
∠HLK
= 180° - 66°
= 114° (Angles on a straight line)
(b)
∠NLM
= 66° - 31°
= 35°
∠LNM
= (180° - 35°) ÷ 2
= 72.5 ° (Isosceles triangle)
Answer(s): (a) 114°; (b) 72.5°