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Since BECF is a rhombus, all sides are equal length.

Given BC = BF, then ΔBEC and ΔBCF are both equilateral triangles, hence the blue 60º labelled in the drawing above.

AC is the diagonal of the square, therefore ∠ACB = 45º

(a) ∠x = 60º – 45º = 15º

∠HEC = 60º / 2 = 30º

∠CHE = 180º – 30º – 15º = 135º

∠GHF = ∠CHE = 135º

(b) ∠y = 180º – 135º – 40º =

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(a)
BC = BF = BE = CE (triangle is an equilateral triangle)
angle BEC = angle BCE = 60 degrees
angle x = 60 – 45 = 15 degrees

(b)
angle CHE = 180 – 15 – 30 = 135 degrees = angle AHF
angle y = 180 – 135 – 40 = 5 degrees

Ans : (a) 15 degrees; (b) 5 degrees.

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