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Unformatted text preview: revious problem. Let t = arccsc(−5) and let
θ = t radians. Then csc(θ) = −5 which means π ≤ θ < 32 . Let α be the reference angle for θ.
Then 0 < α < 2 and csc(α) = 5. Hence, α = arccsc(5) = arcsin 1 radians, where the last
equality comes from Theorem 10.29. Since, in this case, θ = π + α = π + arcsin 1 ≈ 3.3430
radians, we get arccsc(−5) ≈ 3.3430.
y 1 θ α 1 x The inverse trigonometric functions are typically found in applications whenever the measure of an
angle is required. One such scenario is presented in the following example.
Example 10.6.6. 6 The roof on the house below has a ‘6/12 pitch.’ This means that when viewed
from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination
6 The authors would like to thank Dan Stitz for this problem and associated graphics. 716 Foundations of Trigonometry from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded
to the nearest hundredth of a degree. Front View Side View Solution. If...
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