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Question: In the given figure, O is the centre of the circle. If \(\angle PBC = 25^\circ and \angle APB = 110^...

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

Explanation

Solution

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,
PBC=25andAPB=110\angle PBC = 25^\circ and \angle APB = 110^\circ ……………………………. (1)
To find the aADB\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,
ADB=ACB\angle ADB = \angle ACB…………………………………. (2)
Now let’s findACB\angle ACB,
As we all know APC\angle APC is a straight angle,
APC=180\angle APC = 180^\circ
But, APC\angle APCcan be written as,
APC=APB+CPB\angle APC = \angle APB + \angle CPB
180=110+CPB\therefore 180^\circ = 110^\circ + \angle CPB………………………….. [From (1)]
CPB=180110\therefore \angle CPB = 180^\circ - 110^\circ
CPB=70\therefore \angle CPB = 70^\circ ……………………………….. (3)

Now consider BPC\triangle BPC,
As the property of a triangle says that the sum of three angles of a triangle are 180180^\circ ,
CPB+PBC+BCP=180\angle CPB + \angle PBC + \angle BCP = 180^\circ
70+25+BCP=180\therefore 70^\circ + 25^\circ + \angle BCP = 180^\circ
BCP=18095\therefore \angle BCP = 180^\circ - 95^\circ
BCP=85\therefore \angle BCP = 85^\circ
We can write BCP\angle BCP as PCB\angle PCB
PCB=85\therefore \angle PCB = 85^\circ
Now if we see the figure we will come to know that PCB=ACB\angle PCB = \angle ACB as P and A lie on the same line.
ACB=85\therefore \angle ACB = 85^\circ
Our target is to find the ∠ADB therefore rewrite the equation (2)
ADB=ACB\therefore \angle ADB = \angle ACB
Put the value ofACB=85\angle ACB = 85^\circ ,
ADB=85\therefore \angle ADB = 85^\circ
Therefore the value of ADB\angle ADB is 8585^\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.