The figure is not drawn to scale. CFHM is a parallelogram. EJL and DGK are triangles. ∠EJK = 80° and KDG = 52°. Find
- ∠DEJ
- the sum of ∠DKG, ∠EAG and ∠GBJ.
(a)
∠DEJ
= 180° - 80°
= 100° (Interior angles, CF//MH)
(b)
∠EAG = ∠DAB (Vertically opposite angles)
∠GBJ = ∠KBA (Vertically opposite angles)
DKBA is a quadrilateral.
∠DKG + ∠KBA + ∠DAB + 52° = 360° (Sum of angles in a quadrilateral)
∠DKG + ∠KBA + ∠DAB
= 360° - 52°
= 308°
Answer(s): (a) 100°; (b) 308°