Question

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 $\angle ADB$.

Answer 1

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

In $△PCB$

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 ${85}^o$