Question

If the figure, ABCD is a square of side 1 cm while AEF is an equilateral triangle of side 1 cm. The lines BE and DF intersect at G. If B, A, F are collinear points, then find (in degree measure)

Answer 1

Answer:

Given, ABCD is a square. DCE is an equilateral triangle.

ABCD is a square,

AB=BC=CD=DA

DCE is an equilateral triangle,

DC=DC=CE

Hence, AB=BC=CD=DA=CE=DC

Now, In △ADE,

AD=DE

Thus, ∠AED=∠DAE=x

∠ADE=∠ADC+∠EDC

∠ADE=90+60

∠ADE=150

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Sum of angles of triangle ADE = 180

∠ADE+∠AED+∠DAE=180

150+x+x=180

x=15

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Hence, ∠DAE=15

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Step-by-step explanation:

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