In Fig.10.14, \angleACB = 40º. Find \angleOAB.

Answers (1)

50°

Solution:

Given, \angleACB = 40°

As we know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle

\\\Rightarrow \angle AOB = 2 \angle ACB.\\ \\ \Rightarrow \angle ACB = \frac{ \angle AOB}{2}\\ \\ \Rightarrow 40^{\circ} =\frac{1}{2} \angle AOB\\ \\ \Rightarrow \angle AOB = 80^{\circ}

In \triangleAOB, AO = BO (Both are the radius of the same circle)

\angleOBA = \angleOAB (angles opposite to equal sides in a triangle are equal)

As we know, the sum of all three angles in a triangle AOB is 180°.

In \triangleAOB
\\\Rightarrow \angle AOB + \angle OBA + \angle OAB = 180^{\circ}\\ \Rightarrow 80° + \angle OAB + \angle OAB = 180^{\circ}\\ \Rightarrow 2\angle OAB = 180^{\circ} - 80^{\circ}\\ \Rightarrow 2\angle OAB = 100°\\\\ \Rightarrow \angle OAB = \frac{100^{\circ}}{2}=50^{\circ}

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