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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