# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 322: 48

The distance between the battleship and the submarine is 1451.95 feet

#### Work Step by Step

Let the airplane be at point A. Let the battleship be at point B. Let the submarine be at point C. The points ABC form a triangle. Angle A = $24^{\circ}10'-17^{\circ}30' = 6^{\circ}40'$ Angle B = $17^{\circ}30'$ We can find angle C: $C = 180^{\circ}-A-B$ $C = 180^{\circ}-6^{\circ}40'-17^{\circ}30'$ $C = 155^{\circ}50'$ We can use the law of sines to find the distance $BC$, which is the distance between the battleship and the submarine: $\frac{BC}{sin~6^{\circ}40'} = \frac{5120}{sin~155^{\circ}50'}$ $BC = \frac{5120~sin~6^{\circ}40'}{sin~155^{\circ}50'}$ $BC = 1451.95~ft$ The distance between the battleship and the submarine is 1451.95 feet

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