Altitudes AD and BE of acute triangle ABC intersect at point H. If angle ABH = 30 and angle DAC = 45, then what is angle HCA in degrees?
In order to solve this problem, let's analyze somethings about it.
First, let's note that right triangles \(ABC, ABE, ABD, BHD, BEC\) are all 30-60-90 triangles.
Also, note that right triangles \( ABE,ADC \) are 45-45-90 triangles.
In order to solve the problem, let's draw a line to help find HCA for us.
Let's draw the altitude of ABC by connecting c to line segment AB.
This creates a right triangle!
Since we know that
\(\angle BAC =60^\circ\)
HCA would have to be 30 degrees because of the right triangle.
So we have
\(\angle HCA = 30^\circ\)
So 30 degrees is our answer.
Thanks! :)