## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 305: 64a

#### Answer

The ground coordinates of the house are $~~(1131.8, 4390.2)$ The ground coordinates of the fire are $~~(2277.5, -2596.2)$

#### Work Step by Step

We can find the X-coordinate of the house: $X = \frac{(a-h)~x}{f~sec~\theta-y~sin~\theta}$ $X = \frac{(7400-150)~(0.9)}{(6)~sec~4.1^{\circ}-(3.5)~sin~4.1^{\circ}}$ $X = 1131.8$ We can find the Y-coordinate of the house: $Y = \frac{(a-h)~y~cos~\theta}{f~sec~\theta-y~sin~\theta}$ $Y = \frac{(7400-150)~(3.5)~cos~4.1^{\circ}}{(6)~sec~4.1^{\circ}-(3.5)~sin~4.1^{\circ}}$ $Y = 4390.2$ The ground coordinates of the house are $~~(1131.8, 4390.2)$ We can find the X-coordinate of the fire: $X = \frac{(a-h)~x}{f~sec~\theta-y~sin~\theta}$ $X = \frac{(7400-690)~(2.1)}{(6)~sec~4.1^{\circ}-(-2.4)~sin~4.1^{\circ}}$ $X = 2277.5$ We can find the Y-coordinate of the fire: $Y = \frac{(a-h)~y~cos~\theta}{f~sec~\theta-y~sin~\theta}$ $Y = \frac{(7400-690)~(-2.4)~cos~4.1^{\circ}}{(6)~sec~4.1^{\circ}-(-2.4)~sin~4.1^{\circ}}$ $Y = -2596.2$ The ground coordinates of the fire are $~~(2277.5, -2596.2)$

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