HLMQ and HKNQ are parallelograms and JPN is an isosceles triangle. Given that ∠JNK = 39°, ∠HQP = 61° and ∠LNM = 46°, find
- ∠KNL,
- ∠LKN.
(a)
∠JNP
= (180° - 22°) ÷ 2
= 79° (Isosceles triangle)
∠KNL
= 180° - 79° - 39° - 46°
= 16° (Angles on a straight line)
(b)
∠LKN
= 180° - 46° -16°
= 118° (Interior angles)
Answer(s): (a) 16°; (b) 118°