In the given figure marc A X B-200- - Find marcc A Y B-A- 140-B- 160-C- 180-D- 200-

The correct option is B 160°m(arc AXB) + nbsp;m(arc AYB) = 360° → nbsp;m(arc AYB) = 360° nbsp;m(arc AXB) → nbsp;m(arc AYB) = 360° 200° = 160° ...

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Byju's Answer

Standard VII

Mathematics

Major Arc, Minor Arc and Semicircular Arc

In the given ...

Question

In the given figure m(arc AXB) = 200°. Find m(arc AYB).

140°

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160°

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180°

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200°

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Solution

The correct option is B 160°
m(arc AXB) + m(arc AYB) = 360°
$\to m (a r c A Y B) = 360 ° - m (a r c A X B)$
$\to m (a r c A Y B) = 360 ° - 200 °$
$= 160 °$

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In the given figure m(arc AXB) = 200°. Find m(arc AYB).

The correct option is B 160°m(arc AXB) + m(arc AYB) = 360° →m(arcAYB)=360°−m(arcAXB) →m(arcAYB)=360°−200° =160°

The correct option is B 160°
m(arc AXB) + m(arc AYB) = 360°
$\to m (a r c A Y B) = 360 ° - m (a r c A X B)$
$\to m (a r c A Y B) = 360 ° - 200 °$
$= 160 °$