Question

Frank and Marie set sail from the same point. Frank is sailing in the direction S7∘E. Marie is sailing in the direction S13∘W. After 5 hours, Marie was 16 miles due west of Frank. How far had Marie sailed?
Round your answer to four decimal places.

Answer 1

Answer on Question #76495 – Math – Trigonometry

Question

Frank and Marie set sail from the same point. Frank is sailing in the direction S7°E. Marie is sailing in the direction S13°W. After 5 hours, Marie was 16 miles due west of Frank. How far had Marie sailed?

Round your answer to four decimal places.

Solution

1. Frank is sailing in the direction South-East. Marie is sailing in the direction South-West.

2. We will execute the drawing. Denote the final point of Marie by the letter M. Denote the final point of Frank by the letter F. From the points M and F we draw perpendiculars on the axis WE (points M1 and F1). Through point F1 we draw a segment parallel to MF and denote the point P.

3. The required segment in the figure OM is denoted by $x$ (How far had Marie sailed). OF is denoted by $y$. M1F1=16 miles. OM1 is denoted by $a$, then OF1 is 16-a. If FF1 is denoted by $n$, then MP also $n$, PM1 is unknown and equal $m$.

4. In the triangle OMM1, $\sin \angle M1OM = \frac{MM1}{OM}$; $\sin 13{}^{\circ} = \frac{m + n}{x}$;

5. In the triangle OFF1, $\sin \angle F1OF = \frac{FF1}{OF}$; $\sin 7{}^{\circ} = \frac{n}{y}$;

6. In the triangle OFM, by the cosine theorem

$$M F ^ {2} = O M ^ {2} + O F ^ {2} - 2 * O M * O F \cos \angle M O F;$$ $$\angle M O F = 1 8 0 {}^ {\circ} - 1 3 {}^ {\circ} - 7 {}^ {\circ} = 1 6 0 {}^ {\circ};$$ $$M F ^ {2} = x ^ {2} + y ^ {2} - 2 x y \cos \angle M O F.$$

7. In the triangle M1PF1 by the Pythagorean theorem

$$P F 1 ^ {2} = M 1 F 1 ^ {2} + M M 1 ^ {2};$$ $$P F 1 ^ {2} = 1 6 ^ {2} + m ^ {2}$$

$PF1 = MF$ since we built a parallelogram, then $PF1^2 = MF^2 = 16^2 + m^2$

8. Let us formulate the system of equations.

$$\left\{ \begin{array}{c} 1 6 - x \cos 1 3 {}^ {\circ} = y \cos 7 {}^ {\circ}; \\ \sin 1 3 {}^ {\circ} = \frac {m + a}{x}; \\ \sin 7 {}^ {\circ} = \frac {a}{y}; \\ 1 6 ^ {2} + m ^ {2} = x ^ {2} + y ^ {2} - 2 x y \cos 1 6 0 {}^ {\circ}. \end{array} \right.$$

9. Calculate the values by means of the calculator:

$$\cos 1 3 {}^ {\circ} = 0. 9 7 4 3 7; \cos 7 {}^ {\circ} = 0. 9 9 2 5 5; \sin 1 3 {}^ {\circ} = 0. 2 2 4 9 5; \sin 7 {}^ {\circ} = 0. 1 2 1 8 7;$$ $$\cos 1 6 0 {}^ {\circ} = - 0. 9 3 9 6 9.$$ $$\left\{ \begin{array}{c} 1 6 - 0. 9 7 4 4 x = 0. 9 9 2 5 5 y; \\ m + n = 0. 2 2 4 9 5 x; \\ n = 0. 1 2 1 8 7 y; \\ 2 5 6 + m ^ {2} = x ^ {2} + y ^ {2} + 1. 8 7 9 3 8 x y. \end{array} \right.$$

10. From the second equation we substitute $n$ into the third equation and express $m$. From the first equation we find $y$. We substitute all expressions into the fourth equation.

$$\left\{ \begin{array}{c} n = 0.12187y; \\ m = 0.22495 \left(\frac{16 - 0.97437x}{0.99255}\right) - 0.12187y; \\ y = \frac{16 - 0.97437x}{0.99255}; \\ 256 + (-0.12272)^2 (16 - 0.97437x)^2 = x^2 + \left(\frac{16 - 0.97437x}{0.99255}\right)^2 + 1.89349x (16 - 0.97437x). \end{array} \right.$$

The last equation can be transformed and solved with respect to $x$.

11.

$$\begin{array}{l} 256 + 3.85792 - 0.46988x + 0.01431x^2 - x^2 - 257.92151 + 31.41387x - 0.95653x^2 \\ - 30.29578x + 1.84496x^2 = 0; \\ - 0.09726x^2 + 0.64821x + 1.93641 = 0; \\ D = (0.64812)^2 + 4 * 0.09726 * 1.93641 = 1.17340; \\ x_{1,2} = \frac{-0.64812 \pm \sqrt{1.1734}}{-2 * 0.09726} = \frac{0.64812 \mp 1.0832}{0.19452}; \\ x_1 = 8.9007; x_2 = -2.2367. \end{array}$$

The correct value is $x > 0$ because it is a distance, then $x_1 = 8.9007$.

Answer: 8.9007 miles.

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