Question

Altitudes AD and BE of acute triangle ABC intersect at point H. If angle AHB = 123 degrees and angle BAH = 26 degrees, then what is angle HCA in degrees?
Asymptote code below
[asy]
size(150); defaultpen(linewidth(0.8));
pair B = (0,0), C = (3,0), A = (1.1,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);
draw(A--B--C--cycle);
draw(A--P^^B--Q);
label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",P,S); label("$E$",Q,NE); label("$H$",H,NW);
[/asy]

Answer 1

Angle HCA is 26 degrees.

Answer 2

First, I'm sorry but there is a message board for a reason. Please use it, students.Secondly, this answer is wrong.

Answer 3

31

Answer 4

We would appreciate it if you worked out the problems by yourself instead of having another person do it for you. If you have a question, you may use the message board to ask us but it is best if you do not use the internet.

Have fun solving math problems, and you can always ask for help!
🙂

Answer 5

AOPS PROE.

Answer 6

The altitudes of a triangle intersect at the orthocenter of the triangle. So, when we extend $\overline{CH}$ to point $F$ on $\overline{AB}$, we have $\overline{CF}\perp\overline{AB}$ as shown below.

[asy]
size(150); defaultpen(linewidth(0.8));
pair B = (0,0), C = (3,0), A = (1.1,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);
draw(A--B--C--cycle);
draw(A--P^^B--Q);
label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",P,S); label("$E$",Q,NE); label("$H$",H+(0,-0.23),SW);
pair F = foot(C,A,B);
draw(C--F);
label("$F$",F,NW);
draw(rightanglemark(C,P,H,3.5));
draw(rightanglemark(H,F,A,3.5));
draw(rightanglemark(H,Q,C,3.5));
[/asy]

From $\triangle HBA$, we have $\angle HBA = 180^\circ - 123^\circ - 26^\circ = 31^\circ$, and from right triangle $BEA$, we have
\[\angle HBA = \angle ABE = 90^\circ - \angle BAE=90^\circ - \angle BAC.\]
But from right triangle $CFA$, we have
\[\angle HCA = \angle FCA = 90^\circ - \angle FAC = 90^\circ - \angle BAC,\]which matches our expression for $\angle HBA$. So, we have $\angle HCA = \angle HBA = \boxed{31^\circ}$.

Answer 7

nooooooooooooooooooooooooooooooooooooo

Answer 8

Please, there's a message board for a reason