HLMQ and HKNQ are parallelograms and JPN is an isosceles triangle. Given that ∠JNK = 37°, ∠HQP = 62° and ∠LNM = 47°, find
- ∠KNL,
- ∠LKN.
(a)
∠JNP
= (180° - 25°) ÷ 2
= 77.5° (Isosceles triangle)
∠KNL
= 180° - 77.5° - 37° - 47°
= 18.5° (Angles on a straight line)
(b)
∠LKN
= 180° - 47° -18.5°
= 114.5° (Interior angles)
Answer(s): (a) 18.5°; (b) 114.5°