On a semi circle with $AB$ as diameter, a point $C$ is taken so that $m (\angle CAB) = 30^o$. Find $m (\angle ACB)$ and $m (\angle ABC)$.

Given:

On a semi circle with $AB$ as diameter, a point $C$ is taken so that $m (\angle CAB) = 30^o$.

To do:

We have to find $m (\angle ACB)$ and $m (\angle ABC)$.

Solution:

$\angle CAB = 30^o$

$\angle ACB = 90^o$                   (Angle in a semi circle)

$\angle CAB + \angle ACB + \angle ABC = 180^o$

$30^o + 90^o + \angle ABC = 180^o$

$120^o + \angle ABC = 180^o$

$\angle ABC = 180^o - 120^o = 60^o$

Hence, $m \angle ACB = 90^o$ and $m \angle ABC = 60^o$.

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Updated on: 10-Oct-2022

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