if anyone knows the answer plz comment...
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angle AOB = 60°[ABO is an equilateral ∆]
angle AOB = 2 ( angle APB) [Angle subtended by an arc at the centre is double the angle subtended at by any point on the remaining part of the circle]
Therefore, angle APB = 30°
reflex angle AOB = 2 ( angle AQB) [Degree measure theorem]
Therefore, angle AQB = 1/2 reflex angle AOB
Angle AQB = 1/2*300
Therefore, angle AQB = 150°
angle AOB = 2 ( angle APB) [Angle subtended by an arc at the centre is double the angle subtended at by any point on the remaining part of the circle]
Therefore, angle APB = 30°
reflex angle AOB = 2 ( angle AQB) [Degree measure theorem]
Therefore, angle AQB = 1/2 reflex angle AOB
Angle AQB = 1/2*300
Therefore, angle AQB = 150°
jaya12334:
this is not a sum of triangle... this is about circle
sry bt I think this is not the correct answer
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