Mathematics

In the given figure, $$O$$ is the centre of the circle. If $$\angle PBC=25^o$$ and $$\angle APB=110^o$$, find the value of $$\angle ADB$$.


SOLUTION
We know that $$\angle ACB=\angle PCB$$

In $$\triangle PCB$$

Using the angle sum property 

$$\angle PCB+\angle BPC+\angle PBC=180^o$$

We know that $$\angle APB$$ and $$\angle BPC$$ are linear pair

By substituting the values

$$\angle PCB +(180^o -110^o)+25^o =180^o$$

On further calculation 

$$\angle PBC +70^o+25^o =180^o$$

$$\angle PCB +95^o =180^o$$

By subtraction

$$\angle PCB =180^o -95^o$$

So we get

$$\angle PCB =85^o$$

We know that the angles in the same segment of a circle equal

$$\angle ADB=\angle ACB =85^o$$

Therefore, the value of $$\angle ADB$$ is $$85^o$$
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