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°


Consider, the diagram given in question.

Join, OA and OB


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


As, Δ OAC and Δ OBC are isosceles

ACO = CAO and OBC = BCO …..(3)

Also, by case1 we get,




(from (2) and (3))

(from (4))

= 90°

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