Answer:
[tex]\angle LKN=\arctan1.8\sqrt{97}\approx 86.77^{\circ}[/tex]
Step-by-step explanation:
Draw the second altitude MB (see attached diagram).
Quadrilateral HLMB is a rectangle, then LM = HB = 3 units.
Trapezoid KLMN is isosceles trapezoid (because KL=MN), thus
[tex]KH = BN = \dfrac{KN-HB}{2}=\dfrac{13-3}{2}=5\ units\\ \\HN=HB+BN=3+5=8\ units[/tex]
Triangle LHN is right triangle, then by Pythagorean theorem,
[tex]LN^2=LH^2+HN^2\\ \\89^2=LH^2+8^2\\ \\LH^2=89^2-8^2\\ \\LH^2=(89-8)(89+8)\\ \\LH^2=81\cdot 97\\ \\LH=9\sqrt{97}\ units[/tex]
Consider right triangle KLH. In this triangle,
[tex]\tan\angle LKH=\{\tan\angle LKH\}=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{LH}{KH}=\dfrac{9\sqrt{97}}{5}=1.8\sqrt{97}\ units[/tex]
So,
[tex]\angle LKN=\arctan1.8\sqrt{97}\approx 86.77^{\circ}[/tex]
