Altitudes $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ intersect at $H$. If $\angle BAC = 54^\circ$ and $\angle ABC = 52^\circ$, then what is $\angle AHB$?
the figure doesn not show any other angles if u draw it out
am i missing somehting??
+2862
Note that all angles in a triangle sum to 180.
Notice \(\Delta ABE\), Angle BEA is 90, angle A is 54, so angle ABE is 180 - 90 - 54 = 36.
Notice \(\Delta BDA\), Angle HDB is 90, angle B is 52, so angle DAB is 180 - 90 - 52 = 38
Notice \(\Delta ABH\), Angle ABE is 36, angle DAB is 38, so angle AHB is 180 - 36 - 38 = 106
