Answer:
[tex]\sin L = 0.60[/tex]
[tex]tan\ N = 1.33[/tex]
[tex]\cos L = 0.80[/tex]
[tex]\sin N = 0.80[/tex]
Step-by-step explanation:
Given
See attachment
From the attachment, we have:
[tex]MN = 6[/tex]
[tex]LN = 10[/tex]
First, we need to calculate length LM,
Using Pythagoras theorem:
[tex]LN^2 = MN^2 + LM^2[/tex]
[tex]10^2 = 6^2 + LM^2[/tex]
[tex]100 = 36 + LM^2[/tex]
Collect Like Terms
[tex]LM^2 = 100 - 36[/tex]
[tex]LM^2 = 64[/tex]
[tex]LM = 8[/tex]
Solving (a): [tex]\sin L[/tex]
[tex]\sin L = \frac{Opposite}{Hypotenuse}[/tex]
[tex]\sin L = \frac{MN}{LN}[/tex]
Substitute values for MN and LN
[tex]\sin L = \frac{6}{10}[/tex]
[tex]\sin L = 0.60[/tex]
Solving (b): [tex]tan\ N[/tex]
[tex]tan\ N = \frac{Opposite}{Adjacent}[/tex]
[tex]tan\ N = \frac{LM}{MN}[/tex]
Substitute values for LM and MN
[tex]tan\ N = \frac{8}{6}[/tex]
[tex]tan\ N = 1.33[/tex]
Solving (c): [tex]\cos L[/tex]
[tex]\cos L = \frac{Adjacent}{Hypotenuse}[/tex]
[tex]\cos L = \frac{LM}{LN}[/tex]
Substitute values for LN and LM
[tex]\cos L = \frac{8}{10}[/tex]
[tex]\cos L = 0.80[/tex]
Solving (d): [tex]\sin N[/tex]
[tex]\sin N = \frac{Opposite}{Hypotenuse}[/tex]
[tex]\sin N = \frac{LM}{LN}[/tex]
Substitute values for LM and LN
[tex]\sin N = \frac{8}{10}[/tex]
[tex]\sin N = 0.80[/tex]
