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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 5

Solve the right triangle shown in the figure. Round lengths to two decimal places and express angles to the nearest tenth of a degree. B = 16.8°, b = 30.5
Right triangle ABC with angle B 16.8°, side b 30.5, right angle at C, sides a and c labeled.

Guida verificata passo dopo passo
1
Identify the given information: angle B (which corresponds to angle Q) is 16.8° and side b (which corresponds to side q) is 30.5 units.
Since triangle QRP is a right triangle with the right angle at R, use the fact that the sum of angles in a triangle is 180°. Calculate angle P as \(P = 90^\circ - B = 90^\circ - 16.8^\circ\).
Use the sine function to find side p (opposite to angle B): \(\sin(B) = \frac{p}{r}\), but since we don't know r yet, use the cosine function with side q: \(\cos(B) = \frac{q}{r}\), rearranged to find \(r = \frac{q}{\cos(B)}\).
Once you find r, use the sine function to find p: \(p = r \sin(B)\).
Finally, verify your results by checking the Pythagorean theorem: \(p^2 + q^2 = r^2\). Round all lengths to two decimal places and angles to the nearest tenth of a degree.

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