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Question

In the figure, OAB=30o&OCB=57o.AOC=?&BOC=?

78032.jpg

A
AOC=120o&BOC=66o.
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B
AOC=54o&BOC=57o.
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C
AOC=30o&BOC=57o.
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D
AOC=54o&BOC=66o.
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Solution

The correct option is D AOC=54o&BOC=66o.
In the figure given, ΔOAB is an isosceles triangle since two of its sides OA,OB are equal, both being the radii of the circle.
Given OAB=30o, we get OBA=30o.
Also, since the sum of all the angles in a triangle is 180o,AOB=120o
Now, ΔOBC is also an isosceles triangle since OB,OC are again the radii.
OCB=OBC=57o
Since the sum of al the angles in a triangle is 180o,BOC=180o57o57o=66o.
AOB=AOC+BOC
AOC=120o66o=54o

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