2. ## Re: triangle altitude help

Hey vegasgunner.

Your best bet is to draw a diagram (like you did in your other question that you posted) as a starting point.

3. ## Re: triangle altitude help

Hello, vegasgunner!

$\text{Altitudes }AX\text{ and }BY\text{ of acute triangle }ABC\text{ intersect at }H.$
$\text{If }\angle BAC = 43^o\text{ and }\angle ABC = 67^o,\text{ then what is }\angle HCA\,?$

Code:
              C
o
** *  X
Y * *   o
o  * .   *
*   o       *
*  . *H  .     *
* .   *       .   *
*43o   *           . *
A o   *   o   *   *   *   o B
Z
In a triangle, the three altitudes are concurrent.
Hence, $CHZ$ is an altitude.
. . $\angle CZA = 90^o$

In right triangle $CZA,\;\angle ZAC = 43^o.$

Therefore: $\angle HCA = \angle ZCA = 47^o.$