Question

draw a tangent AB to a circle with centre o having point of contact at the point B​

Answer 1

Answer:

We know that, tangent perpendicular to the radius.

∴ ∠BCO=∠BDO=90

o

--- ( 1 )

In △OBC and △OBD

⇒ ∠BCD=∠BDO=90

o

[ from ( 1 ) ]

⇒ OB=OB [ Common side ]

⇒ OC=OD [ Radius of a circle ]

∴ △OBC≅△OBD [ By R.H.S congruence rule ]

⇒ ∠OBC=∠OBD [ C.P.C.T ]

∴ ∠OBC=∠OBD=60

o

In △OBC,

⇒ cos60

o

=

OB

BC

⇒

2

1

=

OB

BC

⇒ OB=2BC