Using Tangent of angles to solve for side LK
[tex]tan\theta=\frac{Opposite}{Adjacent}[/tex]
Given:
[tex]\begin{gathered} Opposite=LK \\ Adjacent=MK=8 \\ \theta=24^0 \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} tan24=\frac{LK}{8} \\ \therefore LK=8tan24^0=3.56182\approx3.6 \end{gathered}[/tex]
Also, Using Cosine of angles to solve for side LM
[tex]cos\theta=\frac{Adjacent}{Hypotenuse}[/tex]
Given:
[tex]\begin{gathered} Adjacent=8 \\ Hypotenuse=LM \\ \theta=24^0 \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} cos24^0=\frac{8}{LM} \\ \therefore LM=\frac{8}{cos24}=8.75709\approx8.8 \end{gathered}[/tex]
Let us now solve for the missing angle L
[tex]\begin{gathered} 24^0+90^0+\angle L=180^0\text{ \lparen sum of angles in a triangle is }180^0) \\ \angle L=180^0-(24^0+90^0)=66 \end{gathered}[/tex]
Final answers
[tex]\begin{gathered} \angle L=66^0 \\ \angle K=90^0 \\ LK=3.6 \\ LM=8.8 \end{gathered}[/tex]
