The figure is not drawn to scale. It shows a parallelogram HJKL and an isosceles triangle EHL next to it. ∠GLE = 14°. GF and ED are straight lines. Find
- ∠p
- ∠m + ∠n
- ∠n
(a)
∠p
= 180° - ∠GLE
= 180° - 14°
= 166° (Angles on a straight line)
(b)
∠m = ∠HLK (Parallelogram)
∠m + ∠n
= 180° - 14° - 59°
= 107° (Angles on a straight line)
(c)
∠EHL = ∠m (Corresponding angles)
∠EHL = ∠LEH (Isosceles triangle)
∠m
= 180° - (∠m + ∠n)
= 180° - 107°
= 73°
∠n
= 180° - ∠m - ∠m
= 180° - 73° - 73°
= 34° (Angles sum of triangle)
Answer(s): (a) 166°; (b) 107°; (c) 34°
