The figure is not drawn to scale. ∠JHK = 41°, JH = JK and HN//JM. Given that ∠h is
12 of ∠f and ∠h is 4 times of ∠g, find
- ∠h
- ∠e
(a)
h : f = 1 : 2
f = 2 h
g : h = 1 : 4
f = 2 x 4= 8
f : g : h = 8 : 1 : 4
∠JKH = 41° (Isosceles triangle)
∠HKL
= 180° - 41°
= 139° (Angles on a straight line)
1 u + 4 u + 8 u + 139° = 360° (Sum of angles in a quadrilateral)
13 u = 360° - 139° = 221°
1 u = 221 ÷ 13 = 17°
∠h
= 4 x 1 u
= 4 x 17°
= 68°
(b)
∠JKH = ∠JHK = 41° (Isosceles triangle)
∠KPM = ∠KHN = 68° (Corresponding angles, HN//JM)
∠e
= 68° - 41°
= 27° (Exterior angle of a triangle)
Answer(s): (a) 68°; (b) 27°
