Question

. In the following figure, AE and BC are equal
and parallel and the three sides AB, CD and
DE are equal to one another. If angle A is 102°.
Find angles AEC and BCD.
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Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

. In the following figure, AE and BC are equal

and parallel and the three sides AB, CD and

DE are equal to one another. If angle A is 102°.

Find angles AEC and BCD.

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

if we will draw line between E and C i.e. EC, then it will be equal and parallel to AB.

So, ECD will be equilateral triangle and angle will be 60o.

and ABCE will be parallelogram, where AE and BC are equal and parallel and AB and EC are equal and parallel.

angle AEC =180°−102°=78°

∠BCD=∠BCE+∠ECD=102°+60°=162°

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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Answer 2

Answer:

given : AB = CD = DE

and angle A = 102

to find : angle AEC and angle BCD

Angle A + E = 180 [ interior angle sum pair ]

angle AEC = 102 + AEC = 180

AEC = 180 - 102

AEC = 78

now angle BAC + ABC = 180[ interior angle sum pair ]

BAC + ABC = 180

102 + ABC = 180

ABC = 180 - 102

ABC = 78

angle ABC + BCD = 180 [ interior angle sum pair ]

72 + BCD = 180

BCD = 180 - 78 = 102

so, AEC = 78 and BCD = 102