Question

AB is a chord of a circle having centre O.if angle AOB =60 , then prove that the chord AB is of radius length

Answer 1

we have,
isosceles Δ BOA,
so, ∠ BAO = ∠ABO,
now, by ASP of a Δ,
∠BAO+∠ABO+∠AOB=180°
⇒2∠BAO=180°-60°
⇒2∠BAO=120°
⇒∠BAO=60°

∴ all angles in ΔBOA are 60°
⇒ΔBOA is equilateral
∴ OA=OB=AB
⇒AB= radius.                              (∵ OA and OB are radii)
HENCE PROVED.

Answer 2

Here, OA = OB = r [radii of same circle]

⇒ ∠A = ∠B ----- (i)

In ΔOAB,

∠O + ∠A + ∠B = 180°

⇒ 60° + ∠A +∠A  = 180° [using eq. (i)]

⇒ ∠A = 60°  

Thus ∠O = ∠A = ∠B = 60°  

⇒ ΔOAB is an equilateral triangle.

⇒ AB = OA = OB = r

Hence proved.