PLS HELP ME!!!!!!!! 99 POINTS Given: KLMN is a trapezoid, KL = MNLN = 89 , LH- altitudeLM = 3, KN = 13. Find: m∠LKN
Accepted Solution
A:
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]