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 $85^o$ You're just one step away

Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
Enrolled Students 85

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