Mathematics

In the given figure, $O$ is the centre of a circle in which $\angle OAB=20^o$ and $\angle OCB=55^o$. Find $\angle BOC$

SOLUTION
We know that $OB=OC$ which is the radius

The base angles of an isosceles triangle are equal

So we get

$\angle OBC =\angle OCB =55^o$

In $\triangle BOC$

Using the angle sum property

$\angle BOC + \angle OCB +\angle OBC =180^o$

By substituting values

$\angle BOC +55^o +55^o =180^o$

On further calculation

$\angle BOC =180^o -110^o$

So we get

$\angle BOC =70^o$

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Subjective Medium Published on 09th 09, 2020
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Subjects 10
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