Question

AB is a chord of a circle with centre O. if POQ.=2AB, prove that angle AOB = 60 degree

Answer 1

AB is a chord of circle with centre O as shown in figure.
according to question,
PQ passing through centre O. hence, PQ is a diameter of circle { as we know, a line segment passing through centre cut the circle at two points is known as diameter of circle }

so, PQ is diameter of circle and we know, half of diameter of circle is known as radius of circle. here POQ = 2AB , means Length of AB is equal to length of radius.

Let radius of circle is r
then, AB = r
from figure it is clear that OA and OB are the radius of circle .
so, OA = OB = r

now, ∆OAB,
OA = OB = AB = r .
all sides of ∆OAB are same so, ∆OAB is equilateral triangle .
hence, all angles of ∆OAB will be 60°
hence, ∠AOB = 60°

Answer 2

Step-by-step explanation:

AB is a chord of circle with centre O as shown in figure.

according to question,

PQ passing through centre O. hence, PQ is a diameter of circle { as we know, a line segment passing through centre cut the circle at two points is known as diameter of circle }

so, PQ is diameter of circle and we know, half of diameter of circle is known as radius of circle. here POQ = 2AB , means Length of AB is equal to length of radius.

Let radius of circle is r

then, AB = r

from figure it is clear that OA and OB are the radius of circle .

so, OA = OB = r

now, ∆OAB,

OA = OB = AB = r .

all sides of ∆OAB are same so, ∆OAB is equilateral triangle .

hence, all angles of ∆OAB will be 60°

hence, ∠AOB = 60°

hope it helped you....