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# altitudes of triangles

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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 $\angle AHB$?

Nov 17, 2019

#1
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the figure doesn not show any other angles if u draw it out

am i missing somehting??

Nov 17, 2019
#2
+2490
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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

Nov 17, 2019