The quadrangle ABCD in figure is a square.
When the triangle BCE is folded at CE as crease, B overlaps with F.
When you connect E and F, A and C with a line, find the angle of the smaller one among the angles in the corners where EF and AC cross.
Since AC is a square diagonal line, ∠ACB = 45°.
Thus, ∠FCG = 32° x 2 - 45° = 19°.
Moreover, as ∠F = ∠B = 90°, ∠FGC = 180° - (19° + 90°) = 71°
When the triangle BCE is folded at CE as crease, B overlaps with F.
When you connect E and F, A and C with a line, find the angle of the smaller one among the angles in the corners where EF and AC cross.
Answer
71°
Solution
As shown in Fig. 1, when the intersection of EF and AC is set to G, the angle to be found is ∠FGC. ∠BCE = ∠FCE = 32° as it is an angle folded at CE.Since AC is a square diagonal line, ∠ACB = 45°.
Thus, ∠FCG = 32° x 2 - 45° = 19°.
Moreover, as ∠F = ∠B = 90°, ∠FGC = 180° - (19° + 90°) = 71°
