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As shown in the figure, two angle bisectors of \(\triangle ABC\),\(\overline {BE}\) and \(\overline {CF}\), intersect at P. If \(\angle EPF= 111^{\circ}\), what is \(\angle A\) in degrees?

 


 Sep 8, 2019
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\(\text{Let }\angle A = x, \angle PBC = \angle PBF = y, \angle PCB = \angle PCE = z\\ \begin{cases} x + y +z = 111^{\circ}\\ x+2y+2z = 180^{\circ} \end{cases} \implies y + z = 69^{\circ}\\ x + 69^{\circ} = 111^{\circ}\\ x = 42^{\circ}\\ \angle A = 42^{\circ}\)

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 Sep 8, 2019

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