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 = 50° and ∠PLQ = 31°. Find
- ∠HLK
- ∠LNM
(a)
∠HLQ
= (180° - 50°) ÷ 2
= 65° (Isosceles triangle)
∠HLK
= 180° - 65°
= 115° (Angles on a straight line)
(b)
∠NLM
= 65° - 31°
= 34°
∠LNM
= (180° - 34°) ÷ 2
= 73 ° (Isosceles triangle)
Answer(s): (a) 115°; (b) 73°