Q. 55.0( 1 Vote )

In the picture, O

Answer :

Case1: The angle got by joining any point on the larger part of the circle to the end of the chord is half the angle got by joining the centre of a circle to these end.



● Join A to the centre O of the circle. This line splits the angle at A into two parts, x° and y° .


● Here Δ AOB and Δ AOC are isosceles triangle


Thus, ABO = OAB = x° and ACO = OAC = y°



● Let us take BOC = c°


(180° – 2x) + (180° – 2y) + c° = 360°


360° –2 (x+y) +c° = 360°


2(x+y) = c°



Thus,



Consider, the diagram given in question.


Join, OA and OB


In, Δ OAB


Since, OA and OB are radius of a circle


Therefore, OAB = OBA (since, angles opposite to


equal side are equal) ..(1)


By, angle sum property,


OAB + OBA + AOB = 180°


OAB + OAB + AOB = 180° (from (1))


2 OAB = 180° – AOB



….(2)


As, Δ OAC and Δ OBC are isosceles


ACO = CAO and OBC = BCO …..(3)


Also, by case1 we get,


…(4)


Consider,


CAO + ABC = CAO + CBO + OBA


(from (2) and (3))


(from (4))


= 90°


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