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

Answer

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Hint: We can solve this problem by using the concept i.e.

Angles inscribed by same arc on the circumference of circle are always EQUAL

Complete step-by-step answer:

We will write the given first,

$\angle PBC = 25^\circ and \angle APB = 110^\circ $……………………………. (1)

To find the a$\angle ADB$ we should know the key concept given below,

Concept: Angles inscribed by same arc on the circumference of circle are always EQUAL

Therefore we can say Angles inscribed by arc AB are equal. That is,

$\angle ADB = \angle ACB$…………………………………. (2)

Now let’s find$\angle ACB$,

As we all know $\angle APC$ is a straight angle,

$\angle APC = 180^\circ $

But, $\angle APC$can be written as,

$\angle APC = \angle APB + \angle CPB$

$\therefore 180^\circ = 110^\circ + \angle CPB$………………………….. [From (1)]

$\therefore \angle CPB = 180^\circ - 110^\circ $

$\therefore \angle CPB = 70^\circ $……………………………….. (3)

Now consider $\triangle BPC$,

As the property of a triangle says that the sum of three angles of a triangle are $180^\circ $,

$\angle CPB + \angle PBC + \angle BCP = 180^\circ $

$\therefore 70^\circ + 25^\circ + \angle BCP = 180^\circ $

$\therefore \angle BCP = 180^\circ - 95^\circ $

$\therefore \angle BCP = 85^\circ $

We can write $\angle BCP$ as $\angle PCB$

$\therefore \angle PCB = 85^\circ $

Now if we see the figure we will come to know that $\angle PCB = \angle ACB$ as P and A lie on the same line.

$\therefore \angle ACB = 85^\circ $

Our target is to find the ∠ADB therefore rewrite the equation (2)

$\therefore \angle ADB = \angle ACB$

Put the value of$\angle ACB = 85^\circ $,

$\therefore \angle ADB = 85^\circ $

Therefore the value of $\angle ADB$ is $85^\circ $.

Note: Always remember to draw diagrams for this type of problems to avoid confusion. Also, the property of a circle given by “Angles inscribed by the same arc on the circumference of a circle are always EQUAL” is very much important to solve this problem.

Angles inscribed by same arc on the circumference of circle are always EQUAL

Complete step-by-step answer:

We will write the given first,

$\angle PBC = 25^\circ and \angle APB = 110^\circ $……………………………. (1)

To find the a$\angle ADB$ we should know the key concept given below,

Concept: Angles inscribed by same arc on the circumference of circle are always EQUAL

Therefore we can say Angles inscribed by arc AB are equal. That is,

$\angle ADB = \angle ACB$…………………………………. (2)

Now let’s find$\angle ACB$,

As we all know $\angle APC$ is a straight angle,

$\angle APC = 180^\circ $

But, $\angle APC$can be written as,

$\angle APC = \angle APB + \angle CPB$

$\therefore 180^\circ = 110^\circ + \angle CPB$………………………….. [From (1)]

$\therefore \angle CPB = 180^\circ - 110^\circ $

$\therefore \angle CPB = 70^\circ $……………………………….. (3)

Now consider $\triangle BPC$,

As the property of a triangle says that the sum of three angles of a triangle are $180^\circ $,

$\angle CPB + \angle PBC + \angle BCP = 180^\circ $

$\therefore 70^\circ + 25^\circ + \angle BCP = 180^\circ $

$\therefore \angle BCP = 180^\circ - 95^\circ $

$\therefore \angle BCP = 85^\circ $

We can write $\angle BCP$ as $\angle PCB$

$\therefore \angle PCB = 85^\circ $

Now if we see the figure we will come to know that $\angle PCB = \angle ACB$ as P and A lie on the same line.

$\therefore \angle ACB = 85^\circ $

Our target is to find the ∠ADB therefore rewrite the equation (2)

$\therefore \angle ADB = \angle ACB$

Put the value of$\angle ACB = 85^\circ $,

$\therefore \angle ADB = 85^\circ $

Therefore the value of $\angle ADB$ is $85^\circ $.

Note: Always remember to draw diagrams for this type of problems to avoid confusion. Also, the property of a circle given by “Angles inscribed by the same arc on the circumference of a circle are always EQUAL” is very much important to solve this problem.

Last updated date: 28th Sep 2023

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