In the right ∆abc (m∠c = 90°) the acute angles are in the ratio 5: 1, i. e. m∠bac : m∠abc = 5: 1. if ch is the altitude to ab and cl is the angle bisector of ∠acb, find m∠hcl.

The ratio is 5:1, and the two angles make a sum of 90, so m∠B=15, and m∠A=75
CL bisects ∠ACB, and ∠ACB=90, so m∠BCL=half of 90=45
m∠CLA, as an exterior angle of ΔBCL, equals m∠B+m∠BCL=15+45=60
CH is the altitude to AB, so m∠CHL=90
m∠HCL=180-m∠CHL-m∠CLH=180-90-60=30
30 degree is the answer. 

45x is the correct answer