In the given figure m arc AXB -120- - Find m arc AYB A- 340-B- 120-C- 140-D- 240-

The correct option is D 240°m(arc AXB) + nbsp;m(arc AYB) = 360° → nbsp;m(arc AYB) = 360° nbsp;m(arc AXB) → nbsp;m(arc AYB) = 360° 120° = 240° ...

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Major Arc, Minor Arc and Semicircular Arc

In the given ...

Question

In the given figure m(arc AXB) = 120°. Find m(arc AYB)

340°

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120°

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140°

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240°

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