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# Find Angle ANB = Angle AOB = Angle AMB=?

Last updated date: 22nd Feb 2024
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Hint: In order to solve the given question , we will be using the theorem which states that the angle subtended by an arc at the center is double the angle subtended by it on any point on the remaining part of the circle .

Therefore, the angle subtended by the arc AB at the centre is ${180^ \circ }$. Give the centre a name like X.
So, $\angle AXB = {180^ \circ }$ . Using the theorem which states that the angle subtended by an arc at the center is double the angle subtended by it on any point on the remaining part of the circle .
$\angle AXB = 2\angle ANB$
$\angle ANB$is the angle made at any point on the circle .
$\angle AXB = 2\angle ANB \\ \dfrac{{\angle AXB}}{2} = \angle ANB \\ \dfrac{{{{180}^ \circ }}}{2} = \angle ANB \\ \angle ANB = {90^ \circ } \;$
Similarly , it can be said that angle inscribed in a semicircle is a right angle so Angle ANB = Angle AOB = Angle AMB=${90^ \circ }$, because like $\angle ANB = {90^ \circ }$,$\angle AOB,\angle AMB$are also the angles made at any point on the circle .
So, the correct answer is “${90^ \circ }$”.