in a regular pentagon ABCDE , draw a diagonal BE and then find thenmeasures of :angle BAE , angle ABE and angle BED answer needed asnfast as possible

Given: ABCDE is a regular pentagon

⇒ Sum of angles of ABCDE = 180° × (5 – 2) = 540°

⇒ Each angle of ABCDE =

⇒ ∠BAE = 108°

Now in ΔABE

AB = AE

⇒ ∠ABE = ∠AEB

and ∠BAE + ∠ABE + ∠AEB = 180°

⇒ 108° + ∠ABE + ∠ABE = 180°

⇒ 2∠ABE = 180° – 108°

and ∠AED = ∠AEB + ∠BED

⇒ ∠BED = ∠AED – ∠AEB

= 108° – 36°

= 72°

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