In the figure, not drawn to scale, DEFG is a square, FGH is an equilateral triangle, FHJK is a rhombus, and EF = HF. Find
- ∠FKJ
- ∠DHE
(a)
∠GFH = 60°
∠EFH
= 90° - 60°
= 30°
FH = FE
∠FHE
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
∠FKJ
= ∠FHE
= 75° (Rhombus)
(b)
∠FEH = ∠FHE = 75°
∠DEH
= 90° - 75°
= 15°
∠DHE
= 180° - 15° - 15°
= 150° (Isosceles triangle DEH)
Answer(s): (a) 75°; (b) 150°