The figure is not drawn to scale. ∠JHK = 41°, JH = JK and HN//JM. Given that ∠f is
13 of ∠d and ∠f is 4 times of ∠e, find
- ∠f
- ∠c
(a)
f : d = 1 : 3
d = 3 f
e : f = 1 : 4
d = 3 x 4= 12
d : e : f = 12 : 1 : 4
∠JKH = 41° (Isosceles triangle)
∠HKL
= 180° - 41°
= 139° (Angles on a straight line)
1 u + 4 u + 12 u + 139° = 360° (Sum of angles in a quadrilateral)
17 u = 360° - 139° = 221°
1 u = 221 ÷ 17 = 13°
∠f
= 4 x 1 u
= 4 x 13°
= 52°
(b)
∠JKH = ∠JHK = 41° (Isosceles triangle)
∠KPM = ∠KHN = 52° (Corresponding angles, HN//JM)
∠c
= 52° - 41°
= 11° (Exterior angle of a triangle)
Answer(s): (a) 52°; (b) 11°
