The figure shows a square DEHJ and a rectangle EFGH. ∠FJH = 35°.
- Find ∠ELJ.
- Find ∠DKF.
(a)
∠ELJ
= 35° + 90°
= 125° (Exterior angle of a triangle)
(b)
∠DHJ = 45°
∠HKJ
= 180° - 35° - 45°
= 100° (Angles sum of triangle)
∠DKF
= ∠HKJ
= 100° (Vertically opposite angles)
Answer(s): (a) 125°; (b) 100°