Question

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.

Answer 1

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.