Answer:
Triangle a: 44.7°
Triangle b: 30.5°
Triangle c: 67.4°
Triangle d: 56.6°
Step-by-step explanation:
Recall: SOHCAHTOA
a. Reference angle = [tex] \theta [/tex]
Opposite = 19
Hypotenuse = 27
Apply SOH:
[tex] sin(\theta) = \frac{Opp}{Hyp} [/tex]
[tex] sin(\theta) = \frac{19}{27} [/tex]
[tex] \theta = sin^{-1}(\frac{19}{27}) [/tex]
[tex] \theta = 44.7 degrees [/tex] (nearest tenth)
b. Reference angle = [tex] \theta [/tex]
Opposite = 33
Adjacent = 56
Apply TOA:
[tex] tan(\theta) = \frac{Opp}{Adj} [/tex]
[tex] tan(\theta) = \frac{33}{56} [/tex]
[tex] \theta = tan^{-1}(\frac{33}{56}) [/tex]
[tex] \theta = 30.5 degrees [/tex] (nearest tenth)
c. Reference angle = [tex] \theta [/tex]
Opposite = 12
Adjacent = 5
Apply TOA:
[tex] tan(\theta) = \frac{Opp}{Adj} [/tex]
[tex] tan(\theta) = \frac{12}{5} [/tex]
[tex] \theta = tan^{-1}(\frac{12}{5}) [/tex]
[tex] \theta = 67.4 degrees [/tex] (nearest tenth)
d. Reference angle = [tex] \theta [/tex]
Hypotenuse = 20
Adjacent = 11
Apply CAH:
[tex] cos(\theta) = \frac{Adj}{Hyp} [/tex]
[tex] cos(\theta) = \frac{11}{20} [/tex]
[tex] \theta = cos^{-1}(\frac{11}{20}) [/tex]
[tex] \theta = 56.6 degrees [/tex] (nearest tenth)
