Question

in the given five sided closed figure find angle eab + angle bcd + angle cde angle dear
​

Answer 1

Step-by-step explanation:

In △ABC,

⇒ ∠ABC+∠BCA+∠CAB=180

o

[ Sum of angles of triangle is 180

o

. ] ---( 1 )

In △CDA,

⇒ ∠CDA+∠DAC+∠ACD=180

o

[ Sum of angles of triangle is 180

o

. ] ---- ( 2 )

In △DEA,

⇒ ∠DEA+∠EAD+∠ADE=180

o

[ Sum of angles of triangle is 180

o

. ] --- ( 3 )

⇒ Pentagon ABCDE=△ABC+△CDA+△DEA

From ( 1 ), ( 2 ) and ( 3 )

⇒ Pentagon ABCDE=∠ABC+∠BCA+∠CAB+∠CDA+∠DAC+∠ACD+∠DEA+∠EAD+∠ADE

=180

o

+180

o

+180

o

=540

o

---- ( 4 )

⇒ ∠EAB+∠ABC+∠BCD+∠CDE+∠DEA=Pentagon(ABCDE)

From ( 4 ),

∴ ∠EAB+∠ABC+∠BCD+∠CDE+∠DEA=540

o

.