Answer: [tex]\angle ACB=110^{\circ}[/tex]
Step-by-step explanation:
Here, The perpendicular bisectors of sides AC and BC of ΔABC intersect side AB at points P and Q respectively, and intersect each other in the exterior (outside) of [tex]\triangle ABC[/tex]
Let L is the mid point of side BC while M is the mid point of side AC.(mentioned on below figure).
Then, In triangles QBL and QLC,
QL=QL, ( reflexive)
BL=LC, ( by the property of mid points)
And, [tex]\angle QLB= \angle QLC[/tex] ( right angles)
Therefore [tex]\triangle QBL\cong \triangle QLC[/tex], (SAS)
Thus, [tex]\angle LQB=\angle LQC[/tex] ( by CPCT)
But, [tex]\angle LQB+\angle LQC+\angle CQP = 180^{\circ}[/tex]
Thus, [tex]\angle LQB=59^{\circ}[/tex]
Now, In [tex]\triangle QBL[/tex],
[tex]\angle QBL+\angle BLQ+\angle LQB=180^{\circ}[/tex]
⇒[tex]\angle QBL=31^{\circ}[/tex]=[tex]\angle ABC[/tex]
Similarly, [tex]\triangle PMC\cong \triangle PMA[/tex]
Then [tex]\angle MPC=\angle MPA[/tex] ( by CPCT)
But, [tex]\angle CPQ+\angle MPC+\angle MPA = 180^{\circ}[/tex]
Thus, [tex]\angle MPA=51^{\circ}[/tex]
Now, In [tex]\triangle MPA[/tex],
[tex]\angle MPA+\angle PAM+\angle AMP=180^{\circ}[/tex]
⇒[tex]\angle MAP=39^{\circ}[/tex]=[tex]\angle BAC[/tex]
Again, In [tex]\triangle ABC[/tex],
[tex]\angle ABC+ \angle BAC+\angle ACB = 180^{\circ}[/tex]
[tex]31^{\circ}+39^{\circ}+\angle ACB = 180^{\circ}[/tex]
⇒[tex]\angle ACB=110^{\circ}[/tex]
