The figure is not drawn to scale. EFHJ is a square and DEJ is an equilateral triangle. FGH is an isosceles triangle. DHG is a straight line.
- Find ∠JDH.
- Find ∠FGH.
(a)
∠HJE = 90°
∠DJE = 60° (Equilateral triangle DJB)
∠DJH
= 90° + 60°
= 150°
∠JDH
= (180° - 150°) ÷ 2
= 30° ÷ 2
= 15° (Isosceles triangle)
(b)
∠DHF
= 90° - ∠DHJ;
= 90° - 15°
= 75°
∠GHF
= 180° - ∠DHF
= 180° - 75°
= 105° (Angles on a straight line)
∠FGH
= (180° - ∠GHC) ÷ 2
= (180° - 105°) ÷ 2
= 75° ÷ 2
= 37.5° (Isosceles triangle)
Answer(s): (a) 15°; (b) 37.5°
