+0

Geometry

-1
2
1
+1420

Altitudes $\overline{AD}$ and $\overline{BE}$ of acute triangle $ABC$ intersect at point $H$. If $\angle AHB = 144^\circ$ and $\angle BAH = 66^\circ$, then what is $\angle HCA$ in degrees?

Jun 20, 2024

#1
+1252
+1

Before we solve the problem, let's note something important.

The sum of the angles in a triangle must add up to $$180 ^\circ$$

However, let's take a look at something.

Adding up the two angles we have, we get $$144 + 66 = 210°$$

This means that triangle AHB would have angles adding up to more than 180.

Thus, this is an invalid triangle.

Thanks! :)

~NTS

Jun 21, 2024
edited by NotThatSmart  Jun 21, 2024

#1
+1252
+1

Before we solve the problem, let's note something important.

The sum of the angles in a triangle must add up to $$180 ^\circ$$

However, let's take a look at something.

Adding up the two angles we have, we get $$144 + 66 = 210°$$

This means that triangle AHB would have angles adding up to more than 180.

Thus, this is an invalid triangle.

Thanks! :)

~NTS

NotThatSmart Jun 21, 2024
edited by NotThatSmart  Jun 21, 2024