Question

In figure (1) given below, equilateral triangle EBC surmounts square ABCD. Find angle BED represented by x.​

Answer 1

Given :

• ∆EBC is an equilateral triangle
• ABCD is a square

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To find :

• ∠BED

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Solution :

∆BCE is an equilateral triangle,

Therefore, all angles are equal

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Let the angles be y

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Sum of all angles of a triangle is 180°

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∠EBC + ∠ECB + ∠CEB = 180°

y + y + y = 180°

3y = 180°

y = $\cancel \frac{180°}{3}$

y = 60°

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Therefore,

∠EBC = ∠BCE = ∠CEB = 60°

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EC = CD

Therefore, ∠CED = ∠CDE

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In ∆ECD

sum of all angles of a triangle is 180°

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∠CED + ∠CDE + ∠DCE = 180°

2∠CED + (60° + 90°) = 180° [∠ECB = 60°; ∠DCB = 90°]

2∠CED = 180° - 150°

2∠CED = 30°

∠CED = $\cancel \frac{30°}{2}$

∠CED = 15°

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∠DEC = ∠CEB - ∠DEB

15° = 60° - x

x = 60° - 15°

x = 45°

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Hence, ∠BED(x°) is = 45°

Answer 2

Answer:

∆BCE is an equilateral triangle,

Therefore, all angles are equal

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Let the angles be y

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Sum of all angles of a triangle is 180°

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∠EBC + ∠ECB + ∠CEB = 180°

y + y + y = 180°

3y = 180°

y = 60°

y = 60°

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Therefore,

∠EBC = ∠BCE = ∠CEB = 60°

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EC = CD

Therefore, ∠CED = ∠CDE

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In ∆ECD

sum of all angles of a triangle is 180°

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∠CED + ∠CDE + ∠DCE = 180°

2∠CED + (60° + 90°) = 180° [∠ECB = 60°; ∠DCB = 90°]

2∠CED = 180° - 150°

2∠CED = 30°

∠CED = 15°

∠CED = 15°

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∠DEC = ∠CEB - ∠DEB

15° = 60° - x

x = 60° - 15°

x = 45°

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Hence, ∠BED(x°) is = 45°