Question

Show all work and round to the nearest 10th. Thank you.

Answer 1

Using Tangent of angles to solve for side LK

`tan\theta=(Opposite)/(Adjacent)`

Given:

`\begin{gathered} Opposite=LK \\ Adjacent=MK=8 \\ \theta=24^0 \end{gathered}`

Therefore,

`\begin{gathered} tan24=(LK)/(8) \\ \therefore LK=8tan24^0=3.56182\approx3.6 \end{gathered}`

Also, Using Cosine of angles to solve for side LM

`cos\theta=(Adjacent)/(Hypotenuse)`

Given:

`\begin{gathered} Adjacent=8 \\ Hypotenuse=LM \\ \theta=24^0 \end{gathered}`

Therefore,

`\begin{gathered} cos24^0=(8)/(LM) \\ \therefore LM=(8)/(cos24)=8.75709\approx8.8 \end{gathered}`

Let us now solve for the missing angle L

`\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}`

Final answers

`\begin{gathered} \angle L=66^0 \\ \angle K=90^0 \\ LK=3.6 \\ LM=8.8 \end{gathered}`