Question

ABCD is a cyclic quadrilateral . O is the center of the circle and <AOC =150° then find <CBD​

Answer 1

Step-by-step explanation:

angle AEC=1/2angleAOC

1/2×150

75°

angleAEC+angleABC=180°

75°+X°=180°

x°=180°-75°

105°

now, angleABC+angleCBD=180°

105°+x°=180°

x°=180°-105°

x=75°

answer..........