
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
606.9k+ views
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

