Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 5.2
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Exercise 69 Page 285

Practice makes perfect
a Here we want to relate the opposite leg and the hypotenuse to the reference angle θ. To do that we should use the sine ratio.
sin θ = Opposite/Hypotenuse By substituting the given sides and angle, we get a trigonometric equation relating the side lengths and θ. We have the following equation for θ. sin θ = 6/9 Let's solve using the inverse sine function.
sin θ = 6/9
sin θ = 2/3

sin^(-1)(LHS) = sin^(-1)(RHS)

θ = sin^(-1) 2/3
θ = 0.72972765622...
θ ≈ 0.73
b Here we want to relate the adjacent leg and hypotenuse to the reference angle θ. To do that we should use the cosine ratio.
cos α = Adjacent/Hypotenuse

By substituting the given sides and angle, we get a trigonometric equation relating the side lengths and α.

Let's solve it using the inverse cosine function.
cos α = 5/7

cos^(-1)(LHS) = cos^(-1)(RHS)

α = cos^(-1) 5/7
α = 0.77519337331...
α ≈ 0.78
c Here, we want to relate the opposite and adjacent leg to the reference angle θ. To do that we should use the tangent ratio.

tan β = Opposite/Adjacent By substituting the given sides and angle, we get a trigonometric equation the side lengths and β.

tan β = 7/9

tan^(-1)(LHS) = tan^(-1)(RHS)

β = tan^(-1)7/9
β = 0.66104316885...
β ≈ 0.66