Question

In the Pentagon ABCDE, angle ADE=18°. If BC=CD. Find angle EDC and angle CDB respectively​

Answer 1

Answer:

oin BD, BE

From isosceles △BCD we get

∠CBD=∠CDB=

2

1

(180

∘

−110

∘

)=

2

1

×70

∘

=35

∘

From the cyclic quadrilateral ABDE

∠BDE=180

∘

−∠BAE=180

∘

−120

∘

=60

∘

From the cyclic quadrilateral BCDE

∠BED=180

∘

−110

∘

=70

∘

ArcAB=ArcBC=∠AEB=∠BDC=35

∘

Sum of all the interior angles of a Pentagon =(2×5−4)rt.∠s=6×90

∘

=540

∘

(i) ∠ABC=540

∘

−(110

∘

+35

∘

+60

∘

+70

∘

+35

∘

+120

∘

)=540

∘

−430

∘

=110

∘

(ii) ∠CDE=∠CDB+∠BDE=35

∘

+60

∘

=95

∘

(iii) ∠AED=∠AEB+∠BED=35

∘

+70

∘

=105

∘

(iv) ∠EAD=∠EBD=180

∘

−(70

∘

+60

∘

)=180

∘

−130

∘

=50

∘