Let the measure of angle \( ABC \) be \( x \). Since the angles of triangle \( ABC \) must sum to \( 180^\circ \), we can express the angles as follows:
- Angle \( ABC = x \)
- Angle \( BAC = 5x \)
- Angle \( ACB = 180^\circ - (5x + x) = 180^\circ - 6x \)
Since triangle \( ABC \) is acute, all angles must be less than \( 90^\circ \). Therefore:
- \( x < 90^\circ \)
- \( 5x < 90^\circ \) implies \( x < 18^\circ \)
- \( 180^\circ - 6x < 90^\circ \) implies \( 6x > 90^\circ \) or \( x > 15^\circ \)
Thus, \( x \) is in the range: \[ 15^\circ < x < 18^\circ \]
Next, let's find the exact measures of the angles:
- If \( x = 15^\circ \), then \( BAC = 75^\circ \) and \( ACB = 90^\circ \), which is not acute.
- If \( x = 17^\circ \), then \( BAC = 85^\circ \) and \( ACB = 51^\circ \), which is acute.
Choosing \( x = 17^\circ \) gives us:
- \( \angle ABC = 17^\circ \)
- \( \angle BAC = 85^\circ \)
- \( \angle ACB = 78^\circ \)
Now, we focus on finding angle \( HCL \). Here:
- \( CH \) is the altitude from \( C \) to line \( AB \).
- \( CL \) is the angle bisector from \( C \) to side \( AB \).
To find \( \angle HCL \):
- Since \( CH \) is the altitude, \( \angle HCA = 90^\circ \) and \( \angle HCB = 90^\circ \).
- The angle bisector \( CL \) divides \( \angle ACB \) into two equal angles: \[ \angle ACL = \angle LCB = \frac{1}{2} \times 78^\circ = 39^\circ \]
Now consider triangle \( HCL \):
- \( \angle HCA = 90^\circ \)
- \( \angle ACL = 39^\circ \)
We can find \( \angle HCL \) using the angle sum in triangle \( HCL \): \[ \angle HCL + \angle HCA + \angle ACL = 180^\circ \] \[ \angle HCL + 90^\circ + 39^\circ = 180^\circ \] \[ \angle HCL + 129^\circ = 180^\circ \] \[ \angle HCL = 180^\circ - 129^\circ = 51^\circ \]
Thus, the measure of angle \( HCL \) is \( \boxed{51^\circ} \).
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