HLMQ and HKNQ are parallelograms and JPN is an isosceles triangle. Given that ∠JNK = 38°, ∠HQP = 65° and ∠LNM = 50°, find
- ∠KNL,
- ∠LKN.
(a)
∠JNP
= (180° - 22°) ÷ 2
= 79° (Isosceles triangle)
∠KNL
= 180° - 79° - 38° - 50°
= 13° (Angles on a straight line)
(b)
∠LKN
= 180° - 50° -13°
= 117° (Interior angles)
Answer(s): (a) 13°; (b) 117°