Question

ab is diameter of the circle with the center o and chord cd is equal to radius oc ac and bd produced to meet at p prove that angel apb equal to 60 degree

Answer 1

join bc and od
now
oc=od=cd
cod is an equilateral triangle
angle cod is 60° (1)
also
angle acb =90 (angle in semicircle ) (2)
w.k.t
angle substended by any arc on circles boundry is equal to the half of angle substended by same arc on circle
2angle bcd =angle cod
by equation1
angle cbd =30 (3)
now
angle cbd +angle apb =angle acb ( ext.angle =int.angle in triangle)
by equation2 and 3
30+angle apb =90
angle apb =60

Answer 2

Answer:

Step-by-step explanation:

join bc and od

now

oc=od=cd

cod is an equilateral triangle

angle cod is 60° (1)

also

angle acb =90 (angle in semicircle ) (2)

w.k.t

angle substended by any arc on circles boundry is equal to the half of angle substended by same arc on circle

2angle bcd =angle cod

by equation1

angle cbd =30 (3)

now

angle cbd +angle apb =angle acb ( ext.angle =int.angle in triangle)

by equation2 and 3

30+angle apb =90

angle apb =60