Question

In the figure (not drawn to scale), ∆EFA is a right-angled triangle with ∠EFA = 90° and ∆FGB is an equilateral triangle.

Answer 1

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In the figure(not drawn to scale), EFA is a right-angled triangle with ∠ EFA =90 degree and FGB is an equilateral triangle, find y−2x.

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Given that ΔFGB is an equilateral triangle

∠BGF =∠GBF = ∠GFB = 60º

Find x:

Sum of angles in a triangle is 180º

∠CFG + ∠FGC + ∠GCF = 180

x + 60 + 92= 180

x + 152 = 180

x = 28º

FInd ∠BFC:

Recall that ∠BFG is 60º

∠BFC + ∠CFG = ∠BFG

∠BFC + 28 = 60

∠BFC= 32º

Find y:

∠EFA is a 90º (Given)

∠EFB + ∠BFC = ∠EFA

y + 32 = 90

y = 58º

FInd y - 2x:

We have found that x = 28º and y = 58º

y - 2x = 58 - 2(28)

y - 2x = 58 - 56

y - 2x = 2

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