Respuesta :
The measure of ∠ACB will be 110°
Explanation
According to the diagram below, [tex]DE[/tex] and [tex]DF[/tex] are the perpendicular bisectors of [tex]AC[/tex] and [tex]BC[/tex] respectively and they intersect side [tex]AB[/tex] at points [tex]P[/tex] and [tex]Q[/tex] respectively.
So, [tex]AE=CE[/tex] and [tex]CF= BF[/tex]
Now, according to the [tex]SAS[/tex] postulate, ΔAPE and ΔCPE are congruent each other. Also, ΔCFQ and ΔBFQ are congruent to each other.
That means, ∠PCE = ∠PAE and ∠FCQ = ∠FBQ
As ∠CPQ = 78° , so ∠PCE + ∠PAE = 78° or, ∠PCE = [tex]\frac{78}{2}= 39[/tex]° and as ∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ = [tex]\frac{62}{2}=31[/tex]°
Now, in triangle CPQ, ∠PCQ = 180°-(78° + 62°) = 180° - 140° = 40°
Thus, ∠ACB = ∠PCE + ∠PCQ + ∠FCQ = 39° + 40° + 31° = 110°
[tex]\rm \angle ACB = 110^\circ[/tex]
Step-by-step explanation:
Given :
[tex]\rm m \angle CPQ = 78^\circ[/tex]
[tex]\rm m \angle CQP = 62^\circ[/tex]
The perpendicular bisectors of sides AC and BC of triangle ABC intersect side AB at points P and Q respectively and intersect each other in the exterior of triangle ABC.
AE = CE and CF = BF
Solution :
According to side angle side (SAS) postulate triangle APE and triangle CPE are congruent to each other. Therefore,
[tex]\rm \angle PCE = \angle PAE[/tex] ---- (1)
[tex]\rm \angle FCQ = \angle FBQ[/tex] ---- (2)
Now,
[tex]\rm \angle CPQ = \angle PCE + \angle PAE = 78^\circ[/tex]
[tex]\rm \angle PCE = \dfrac{78^\circ}{2}=39^\circ[/tex] (from (1))
Now,
[tex]\rm \angle CQP = \angle FCQ + \angle FBQ = 62^\circ[/tex]
[tex]\rm \angle FCQ = \dfrac{62^\circ}{2}=31^\circ[/tex] (from(2))
Now in triangle CPQ,
[tex]\rm 78^\circ + 62^\circ + \angle QCP = 180^\circ[/tex]
[tex]\rm \angle QCP =40^\circ[/tex]
Now,
[tex]\rm \angle ACB = \angle QCP + \angle PCE + \angle FCQ = 39^\circ + 31^\circ + 40 ^\circ[/tex]
[tex]\rm \angle ACB = 110^\circ[/tex]
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