Question

Two Tangents AB and AC are drawn to a circle with centre O such that angle BAC=120°.Prove that OA=2AB.

Answer 1

Answer:

OA = 2AB

Step-by-step explanation:

Two Tangents AB and AC are drawn to a circle with centre O such that angle BAC=120°.Prove that OA=2AB.

AB = AC  ( Equal Tangents)

∠BAC = 120°

∠OBA = ∠OCA = 90°

=> ∠BOC = 360° - 120° - 90° - 90° = 60°

in ΔOAB & ΔOAC

AB = AC (tangents)   , OA = OA (common) OB = OC (Radius)

=> ΔOAB ≅  ΔOAC

=> ∠BOA = ∠COA

∠BOA + ∠COA = 60°

=> ∠BOA = 30°

in Δ OAB

∠OBA = 90° & ∠BOA = 30°

=> Sin30 = AB/OA

=> 1/2 = AB/OA

=> OA = 2AB

QED

Proved

Answer 2

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°