Mathematics

# In the given figure, $O$ is the centre of the circle. If $\angle ABD=35^o$ and $\angle BAC=70^o$, find $\angle ACB$.

##### SOLUTION
We know that $BD$ is the diameter of the circle

Angle in a semicircle is a right angle

$\angle BAD=90^o$

Consider $\triangle BAD$

Using the angle sum property

$\angle ADB +\angle BAD +\angle ABD=180^o$

By substituting the values

$\angle ADB +90^o +35^o =180^o$

On further calculation

$\angle ADB =180^o - 90^o -35^o$

By subtraction

$\angle ADB =180^o -125^o$

So we get

$\angle ADB =55^o$

We know that the angle in the same segment of a circle are equal

$\angle ACB =\angle ADB=55^o$

So we get

$\angle ACB =55^o$

Therefore, $\angle ACB =55^o$

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Subjective Medium Published on 09th 09, 2020
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