Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 303: 37

Answer

The initial distance between the ship and the lighthouse is 5.1 miles The final distance between the ship and the lighthouse is 7.2 miles

Work Step by Step

Let $A$ be the first position of the ship. Then angle $A = 180^{\circ} - 37^{\circ}$ which is $143^{\circ}$. Let $C$ be the final position of the ship. Then angle $C = 25^{\circ}$. Let the lighthouse be located at the position of angle $B$. We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-143^{\circ}-25^{\circ}$ $B = 12^{\circ}$ We can find the length of side $c$ which is the initial distance between the ship and the lighthouse: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $c = \frac{b~sin~C}{sin~B}$ $c = \frac{(2.5~mi)~sin~(25^{\circ})}{sin~(12^{\circ})}$ $c = 5.1~mi$ The initial distance between the ship and the lighthouse is 5.1 miles We can find the length of side $a$ which is the final distance between the ship and the lighthouse: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $a = \frac{b~sin~A}{sin~B}$ $a = \frac{(2.5~mi)~sin~(143^{\circ})}{sin~(12^{\circ})}$ $a = 7.2~mi$ The final distance between the ship and the lighthouse is 7.2 miles
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