Question

prove that the interior angle of a regular pentagon is three times the exterior angle of a regular decagon

Answer 1

sum of interior angles of a polygon = (n-2)*180° where n is no. of sides
therefore sum of angles of Pentagon = 3*180° = 540°
measure of 1 interior angle of regular Pentagon = 540°/5 = 108° ___ (i)
sum of interior angles of decagon = 8*180° = 1440°
measure of 1 interior angle of regular decagon = 1440°/10 = 144°
therefore measure of 1 exterior angle of regular decagon = 180°-144° = 36° ___ (ii)
from (i) and (ii)
108° = 3(36°)
(interior angle of Pentagon) = 3(exterior angle of regular decagon)

Answer 2

answer:

for a regular pentagin, no. of sides =n=5

each interior angle of a regular petagon={n-2)*180°/n

=(5-2)*180°/5

=3*180°/5

=36°

fir a decagon no. of sides=n=19

sum of the exterior angles of a regular decagon=360°

each exterior angle of a regular decagon=360°/10=36°

interior angle of a regular pentagon=108°=3*36°

3 times of the exterior angle of a regular decagon