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ABCDE is a regular pentagon. Find each angle of triangleBDE.

Sum of the interior angles of the pentagon ABCDE=180°n-2, where n is the number of sides=180°5-2=180°3=540°Since, ABCDE is a regular pentagon so all interior angles are equal.⇒Each interior angle =540°5=108°Now draw two diagonals joining BE and BD. We  need to find the angles of the ∆BDE.


Consider ∆ABE,AB=AE [regular pentagon]⇒∠AEB=∠ABE [angles opposite to equal sides are equal]⇒∠EAB+2∠AEB=180° [triangle property]⇒108°+2∠AEB=180°⇒∠AEB=36° and ∠ABE=36°We know that ∠AED=108°⇒∠BED=108°-36°=72°Similarly, using ∆BCD we get ∠BDE=72°In ∆BDE,∠BED+∠BDE+∠EBD=180° [triangle property]⇒72°+72°+∠EBD=180°⇒∠EBD=180°-144°=36°Thus, triangle BDE forms an isosceles triangle with angles 72°, 72° and 36°.

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