Mathematics

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

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

The base angles of an isosceles triangle are equal

So we get

$\angle OBA =\angle OAB =20^o$

In $\triangle AOB$

Using the angle sum property

$\angle AOB + \angle OAB +\angle OBA =180^o$

By substituting values

$\angle AOB +20^o +20^o =180^o$

On further calculation

$\angle AOB =180^o -20^o -20^o$

By subtraction

$\angle AOB =180^o -40^o$

So we get

$\angle AOB =140^o$

We know that

$\angle AOC =\angle AOB -\angle BOC$

By substituting the values

$\angle AOC =140^o -70^o$

So we get

$\angle AOC =70^o$

You're just one step away

Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
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