Question

In the given figure <BAC = 30° , show that the chord BC is equal to the radius of circle.​

Answer 1

angle BOC = 2BAC

so, BOC = 60°

BO = OC ( radius of circle )

so,

angle OBC = angle OCB

let angle OBC = x

now,

BOC + OBC + OCB = 180

60 + 2OBC = 180

2OBC = 130

OBC = 60

which means,

OBC = 60

OCB = 60

BOC = 60

Hence triangle is equitarial. ( all side are equal )

so,

OB = OC = BC

BC = radius of circle

Answer 2

given,<BAC=30

To prove:BC=OB=OC

Proof:-<bac=30°

therefore,<boc=2<bac [by theorem]

<boc=60°

in ∆OBC,

ob=oc=radius

<obc=<ocb. [by isosceles∆ property]

<boc+<obc+<ocb=180°

60°+2<obc=180°

2<obc=120°

<obc=60°=<ocb

all angles of ∆OBC are 60°, therefore it is an equilateral triangle.

Therefore,OB=OC=BC

The chord BC is equal to the radius of the circle.

------PROVED------