Question

in the given five sided closed figure find angle EAB + angle ABC+ angle BCD + angle CDE + angleDEA​

Answer 1

Answer:

angle ABC IS 90 ° As triangle ABC is right angled triangle

angle BAC Is equal to angle BCA

so , angle BAC + angle ABC + angle BCA = 180 °

90 + X+ X= 180 °

2X= 180-90

2X=90

X=90÷2

X =45

SO angle BAC ,BCA is 45°

angle CAD= angle BCA ==)45 ( alternate angle )

angle CAD=angle CDA==)45 ( Adjacent angle)

angle CDA=angle DAE ==)45( alternate angle)

angle DAE=angleADE ==)45 ( ADJACENT ANGLE)

So <EAB =45+45==)90°

<ABC=90°

<BCD=45+45==)90°

<CDE=45+45==)90°

<DEA=45+45+X==)180°

=X==)180-90

=X==)90°