H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle of triangle ABC at $A', B' and C'.$ We know angle AHB:angle BHC:angle CHA=9:10:11. Find angle A'B'C' in degrees.
+26367 H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle
of triangle ABC at A', B' and C'.
We know angle AHB:angle BHC:angle CHA=9:10:11.
Find angle A'B'C' in degrees.
\(\begin{array}{|rcll|} \hline \mathbf{\angle AHB:\angle BHC:\angle CHA} &=& \mathbf{9:10:11} \\\\ \frac{\angle AHB}{\angle CHA} &=& \frac{9}{11} \\ \angle AHB &=& \frac{9}{11} *( \angle CHA ) \\\\ \frac{\angle BHC}{\angle CHA} &=& \frac{10}{11} \\ \angle BHC &=& \frac{10}{11}*(\angle CHA ) \\\\ \angle AHB+\angle BHC+\angle CHA &=& 360^\circ \\ \frac {9}{11} *( \angle CHA )+\frac{10}{11}*(\angle CHA )+\angle CHA &=& 360^\circ \\ \frac {19}{11} *( \angle CHA )+\angle CHA &=& 360^\circ \\ \frac {30}{11} *( \angle CHA ) &=& 360^\circ \\ \angle CHA &=& \frac {11}{30} *360^\circ \\ \mathbf{\angle CHA} &=& \mathbf{132^\circ} \\ \hline \end{array}\)
\(\text{$\angle A'B'C'$ in degrees is $2*42^\circ = 84^\circ$}\)
