Question

In the given figure O is the center of the given circle . AB is a side of a square BC is a side of regular Pentagon and cd side of regular hexagon. Find AOB , AOC, AOD and BCD.

Answer 1

As given that AB is the side of pentagon the angle subtended by each arm of the pentagon at the centre of the circle is =

5

360

o

=72

o

Thus angle ∠AOB=72

o

Similarly as BC is the side of a hexagon hence the angle subtended by BC at the centre is =

6

360

o

i.e., 60

o

∠BOC=60

o

The triangle thus formed, △BOC is an isosceles triangle with OB=OC as they are radii of the same circle.

Thus ∠OBC=∠OCB as they are opposite angles of equal sides of an isosceles triangles.

The sum of all the angles of a triangles is 180

so, ∠BOC+∠OBC+∠OCB=180

2∠OBC+60

=180

as, ∠OBC=∠OCB

2∠OBC=180

−60

2∠OBC=120

∠OBC=60

as ∠OBC=∠OCB

So, ∠OBC=∠OCB=60

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