In the figure, FG = FJ, ∠GFJ = 31° and ∠KHJ = 15°. Find
- ∠b
- ∠c
(a)
∠FGJ
= (180° - 31°) ÷ 2
= 74.5° (Isosceles triangle)
∠b
= 180° - 15° - 74.5°
= 90.5° (Angles sum of triangle)
(b)
∠FJH
= 180° - 74.5°
= 105.5° (Angles on a straight line)
∠c
= 15° + 105.5°
= 120.5° (Exterior angle of a triangle)
Answer(s): (a) 90.5°; (b) 120.5°