Maths / Circles / Angle at the Centre is Double The Angle at Circumference

QUESTION

 If  $\angle AOB=110^o,\;then \; \angle ADB+\angle BEC$ is equal to ​​​​​​​
 OPTIONS A. $360^o$ B. $110^o$ C. $220^o$ D. $90^o$
Right Option : D

EXPLANATION
Explain TypeExplanation Content
Text

$\angle AOB=110^o,\;then \; \angle BOC= 180^0-110^0=70^0$ angles on a straight line

Angle at the centre is double the angle at the citrcumference

$\angle ADB=\frac{\angle AOB}{2}= \frac{110^o}{2}= 55^0$

$\angle BEC=\frac{\angle BOC}{2}= \frac{70^o}{2}= 35^0$

$\angle ADB+\angle BEC= 55^0+35^0=90^0$

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