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Ch. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 1

Solve the right triangle shown in the figure. Round lengths to two decimal places and express angles to the nearest tenth of a degree. A = 23.5°, b = 10

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1
Identify the given information: angle \(A = 23.5^\circ\) and side \(b = 10\). From the figure, angle \(A\) corresponds to angle \(Q\), and side \(b\) corresponds to side \(p\) (opposite angle \(Q\)).
Since the triangle is right-angled at \(R\), angle \(P\) can be found using the fact that the sum of angles in a triangle is \(180^\circ\). So, calculate angle \(P\) as \(90^\circ - 23.5^\circ\).
Use the sine function to find the hypotenuse \(r\) because \(\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\sin(23.5^\circ) = \frac{p}{r}\), so rearrange to find \(r = \frac{p}{\sin(23.5^\circ)}\).
Use the cosine function to find side \(q\) (adjacent to angle \(Q\)) because \(\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\). So, \(\cos(23.5^\circ) = \frac{q}{r}\), and rearranged \(q = r \times \cos(23.5^\circ)\).
Calculate all values using the above formulas, rounding lengths to two decimal places and angles to the nearest tenth of a degree as required.

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