Hello, vegasgunner!
$\displaystyle \text{Altitudes }AX\text{ and }BY\text{ of acute triangle }ABC\text{ intersect at }H.$
$\displaystyle \text{If }\angle BAC = 43^o\text{ and }\angle ABC = 67^o,\text{ then what is }\angle HCA\,?$
In a triangle, the three altitudes are concurrent.Code:C o ** * X Y * * o o * . * * o * * . *H . * * . * . * *43^{o} * . * A o * o * * * o B Z
Hence, $\displaystyle CHZ$ is an altitude.
. . $\displaystyle \angle CZA = 90^o$
In right triangle $\displaystyle CZA,\;\angle ZAC = 43^o.$
Therefore: $\displaystyle \angle HCA = \angle ZCA = 47^o.$