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In figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB. |
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Answer» By degree measure theorem: ∠AOB = 2 ∠APB so, ∠AOB = 2 × 70° = 140° Since AOBC is a cyclic quadrilateral, we have ∠ACB + ∠AOB = 180° ∠ACB + 140° = 180° ∠ACB = 40° |
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