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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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