In ∆ABC acute angles are in the ratio 5:1, i.e. ∠BAC : ∠ABC = 5:1. If CH is an altitude and CL is an angle bisector, find m∠HCL.

If [tex]m\angle BAC:m\angle ABC=5:1,[/tex] then [tex]m\angla BAC=(5x)^{\circ},\ m\angle ABC=x^{\circ}.[/tex]The sum of the measures of the interior angles of the triangle is always 180°, then[tex]m\angle ACB=180^{\circ}-x^{\circ}-(5x)^{\circ}=180^{\circ}-(6x)^{\circ}.[/tex]Since CL is bisector, then[tex]m\angle BCL=m\angle ACL=\dfrac{180^{\circ}-(6x)^{\circ}}{2}=90^{\circ}-(3x)^{\circ}.[/tex]Consider right triangle ACH. In this triangle [tex]m\angle ACH=180^{\circ}-90^{\circ}-(5x)^{\circ}=90^{\circ}-(5x)^{\circ}.[/tex]Now angle HCL has measure[tex]m\angle HCL=m\angle ACL-m\angle ACH=90^{\circ}-(3x)^{\circ}-(90^{\circ}-(5x)^{\circ})=(2x)^{\circ}=2m\angle ABC.[/tex]Answer: [tex]m\angle HCL=2m\angle ABC.[/tex]