Question

If for a regular pentagon ABCDE, the lines AD and BE intersect at points P, then ∠BAD and ∠APE respectively are 

(a) 36° and 72° 

(b) 54° and 108° 

(c) 72° and 108° 

(d) 36° and 108°

Answer 1

(c) 72° and 108° 

Sum of the interior angles of a regular pentagon = 540° 

⇒ Each interior angle = \(\frac{540°}{5}\) = 180° 

As AD || BC 

∠BAD + ∠ABC = 180° (co-int. ∠s) 

⇒ ∠BAD = 180° – 108° = 72° 

⇒ ∠PAE = 108° – 72° = 36° 

Also, applying BE || DC, ∠BED = 72° ⇒ ∠PEA = 36° 

∴ ∠APE = 180° – (∠PAE + ∠PEA)       (Angle sum property of a Δ) 

= 180° – (36° + 36°) = 180° – 72° = 108°.