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?
+14995 As shown in the figure, two bisectors of ABC, BE and CF, intersect at P. If EPF = 111 °, what is \(\angle A \) in degrees?
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\(\dfrac{180°-\alpha }{2}+\dfrac{180°-111°}{2}\color{BrickRed}=90°\ |\ mistake!\\ 180°-\alpha +180°-111°=2\cdot 90°\\ \alpha=180° +180°-111°-2\cdot 90° \)
\(\alpha \neq69°\)
\(\frac{\angle B}{2}+\frac{\angle C}{2}+111°=180°\)
\(\frac{\angle B}{2}+\frac{\angle C}{2}=180°-111°\\ (\angle B+\angle C)=2\cdot (180°-111°)\\ (\angle B+\angle C)+\angle A=180°\\ \angle A=180°-(\angle B+\angle C)\\ \angle A=180°-2\cdot (180°-111°)\)
\(\angle A=42° \)
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