Question

in the given figureOis the centre of the circle. if AOB =140 & OAC=50; FIND
1) ACB
2) OBC
3) OAB
4) CBA. PLZZ show full workin g with proper steps and diagram photo..

Answer 1

Take a point D on the left side of O (in major segment) on the circle.
Join AD & BD.

∠ADB = 1/2* ∠AOB = 70° (angle made at circle is half of angle at the center)
ACBD is a cyclic quadrilateral.  Hence, ∠ADB + ∠ACB = 180°.
∠ACB = 110°

In the quadrilateral OACB,   ∠OBC = 360 - 140 -50 - 110 = 60°

ΔOAB is isosceles, as OA = AB.  
     So ∠OAB = ∠OBA = 1/2 (180 - AOB) = 20°

So ∠ABC = 60- 20 = 40°      ans  BAC = 50 - 20 = 30°

Answer 2

Answer:

Take a point on D on the left side of O (in the major segment) on the circle.

Join AD and BD

Angle ADB =1/2,AOB =70 degree ( angle made at circle is half of angle at the center)

ACBD is a cyclic quadrilateral .

Hence, ADB+ACB=180 degree

ACB =100 degree

In the quadrilateral OACB, angle OBC=360-140-50-110 = 60 degree

OAB is an isosceles, as OA=AB .

So angle OAB =OBA=1/2(180-AOB) =20 degree

So angle ABC=60-20=40degree ans BAC=50-20 =30degree..

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