Question

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.

Answer 1

If
`m\angle BAC:m\angle ABC=5:1,` then
`m\angla BAC=(5x)^(\circ),\ m\angle ABC=x^(\circ).`

The sum of the measures of the interior angles of the triangle is always 180°, then

`m\angle ACB=180^(\circ)-x^(\circ)-(5x)^(\circ)=180^(\circ)-(6x)^(\circ).`

Since CL is bisector, then

`m\angle BCL=m\angle ACL=(180^(\circ)-(6x)^(\circ))/(2)=90^(\circ)-(3x)^(\circ).`

Consider right triangle ACH. In this triangle

`m\angle ACH=180^(\circ)-90^(\circ)-(5x)^(\circ)=90^(\circ)-(5x)^(\circ).`

Now angle HCL has measure

`m\angle HCL=m\angle ACL-m\angle ACH=90^(\circ)-(3x)^(\circ)-(90^(\circ)-(5x)^(\circ))=(2x)^(\circ)=2m\angle ABC.`

Answer:
`m\angle HCL=2m\angle ABC.`