Question

Given that DAB and EAC are straight lines find the value of unknown.​

Answer 1

Answer:

triangle ADC is an equilateral triangle

so each angle is 60°

DAB is a straight line hence it has an angle of 180°

angle DAC + angle BAC = 180°

angle BAC = 180° - 60° = 120°

in triangle ACB

angle ACB = angle ABC = 2y°

2y+2y+120=180

4y = 60

y = 15°

unknown angle is 2y = 2×15 = 30°

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Answer 2

  △ DCA is equilateral triangle .....{sides are equal}

•°• ∠CDA = ∠DCA = ∠DAC = 60° ✅

In △ CAB is isosceles triangle ......{two sides are equal}

•°• ∠ACB = ∠ABC = 2y° ....1..{angle opposite to congruent sides are congruent}

now, line DAB is a straight line

•°• ∠DAC + ∠CAB = 180°

∠CAB = 180 - 60

∠CAB = 120° .............2

IN  △ CAB

∠A + ∠C + ∠B = 180° .......{+ of all angles}

120° + 2y° + 2y° = 180° .......{from1 and 2}

4y = 180 - 120

4y = 60

y = 60/4

y = 15✅

The value for 2y° = 2× 15 = 30° ✅