HLMQ and HKNQ are parallelograms and JPN is an isosceles triangle. Given that ∠JNK = 36°, ∠HQP = 62° and ∠LNM = 50°, find
- ∠KNL,
- ∠LKN.
(a)
∠JNP
= (180° - 27°) ÷ 2
= 76.5° (Isosceles triangle)
∠KNL
= 180° - 76.5° - 36° - 50°
= 17.5° (Angles on a straight line)
(b)
∠LKN
= 180° - 50° -17.5°
= 112.5° (Interior angles)
Answer(s): (a) 17.5°; (b) 112.5°