In the figure, ∠ZEJ is a right-angled isosceles triangle. ZJ // BH , ∠GDC = 46°, ∠EGF = 43° and ∠CFD = 54°. Find
- ∠ZJE
- ∠DBH
- ∠FHG
(a)
∠ZJE
= (180° - 90°) ÷ 2
= 90° ÷ 2
= 45° (Isosceles triangle)
(b)
∠EGA = 45° (Corresponding angles, JZ//GA)
∠DGB
= ∠EGF + ∠EGA
= 43° + 45°
= 88°
∠DBH
= 180° - ∠GDC - ∠DGB
= 180° - 46° - 88°
= 46° (Angles sum of triangle)
(c)
∠FGH
= 180° - ∠DGB
= 180° - 88°
= 92°(Angles in a straight line)
∠HFG = ∠DFE = 54° (Vertically opposite angles)
∠FHG
= 180° - ∠HFG - ∠FGH
= 180° - 54° - 92°
= 34° (Angles sum of triangle)
Answer(s): (a) 45°; (b) 46°; (c) 34°