In the figure, ∠YDH is a right-angled isosceles triangle. YH // AG , ∠FCB = 51°, ∠DFE = 43° and ∠BEC = 55°. Find
- ∠YHD
- ∠CAG
- ∠EGF
(a)
∠YHD
= (180° - 90°) ÷ 2
= 90° ÷ 2
= 45° (Isosceles triangle)
(b)
∠DFZ = 45° (Corresponding angles, HY//FZ)
∠CFA
= ∠DFE + ∠DFZ
= 43° + 45°
= 88°
∠CAG
= 180° - ∠FCB - ∠CFA
= 180° - 51° - 88°
= 41° (Angles sum of triangle)
(c)
∠EFG
= 180° - ∠CFA
= 180° - 88°
= 92°(Angles in a straight line)
∠GEF = ∠CED = 55° (Vertically opposite angles)
∠EGF
= 180° - ∠GEF - ∠EFG
= 180° - 55° - 92°
= 33° (Angles sum of triangle)
Answer(s): (a) 45°; (b) 41°; (c) 33°
