In the figure, not drawn to scale, BCDE is a square, DEF is an equilateral triangle, DFGH is a rhombus, and CD = FD. Find
- ∠FGH
- ∠BFC
(a)
∠EDF = 60°
∠CDF
= 90° - 60°
= 30°
DF = DC
∠DFC
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
∠FGH
= 180° - 75°
= 105° (Interior angles)
(b)
∠DCF = ∠DFC = 75°
∠BCF
= 90° - 75°
= 15°
∠BFC
= 180° - 15° - 15°
= 150° (Isosceles triangle)
Answer(s): (a) 105°; (b) 150°