Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 297: 28

Answer

1.93 miles

Work Step by Step

The triangle formed between stations A and B and the Receiver R will be ∆ABR. We know the line AB to be 3.46 miles long, so we have one side of the triangle and are indirectly given two of its angles, A and B. Find angle A by subtracting the 47.7º from the right angle between the East and North. $90º - 47.7º=42.3$ Find angle $B$ by subtracting 270º from the given 302.5º. $302.5º-270º=32.5º$ The remaining angle R can be found because the sum of a triangle's interior angles must be 180º. So, $R+42.3º+32.5º=180º$ Solve for R $R = 180º-42.3º-32.5º$ $R = 105.2º$ Now use the law of sines to relate angle R and line AB with line AR and its opposite, angle B. The law of sines relates the length of a triangle's sides with the sine of their opposite angle, such that: $\frac{AB}{\sin R} = \frac{AR}{\sin B}$ Solve for AR $ {AR}= \frac{{AB}\times \sin B}{\sin R}$ Plug known values in. $ {AR}= \frac{{3.46}\times \sin 32.5}{\sin 105.2}$ Solve with a calculator $AR ≈ 1.93$ The distance between point A and point R is about 1.93 miles
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