# A ship leaves port on a bearing of 34.0° and travels 10.4 mi. The ship then turns due east and travels 4.6 mi. How far is the ship from port, and what is its bearing from port?

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Jun 14, 2017

$13.5 \text{mi}$ at a bearing of ${50.4}^{\circ}$

#### Explanation:

Bearing is a clockwise angle measured from due North. This is a problem, because all of the trigonometric functions are referenced to a counterclockwise angle measured from East.

A bearing of ${34}^{\circ}$ corresponds to a trigonometric angle of ${\theta}_{1} = {90}^{\circ} - {34}^{\circ} = {56}^{\circ}$

The (x,y) values for the position of the ship after completing its first heading are:

$x = \left(10.4 \text{mi}\right) \cos \left({56}^{\circ}\right)$
$y = \left(10.4 \text{mi}\right) \sin \left({56}^{\circ}\right)$

The trigonometric angle for the second heading is ${\theta}_{2} = {90}^{\circ} - {90}^{\circ} = {0}^{\circ}$

The (x,y) values for the position of the ship after completing its second heading is:

x = (10.4"mi")cos(56^@) + (4.6"mi")cos(0^@)~~ 10.4"mi"
y = (10.4"mi")sin(56^@)+ (4.6"mi")sin(0^@) ~~ 8.6"mi"

The distance from port is:

$d = \sqrt{{\left(10.4\right)}^{2} + {\left(8.6\right)}^{2}} \approx 13.5 \text{mi}$

Its trigonometric angle is:

$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

$\theta = {\tan}^{-} 1 \left(\frac{8.6}{10.4}\right)$

$\theta \approx {39.6}^{\circ}$

The bearing angle is:

${\theta}_{b} = {90}^{\circ} - {39.6}^{\circ} = {50.4}^{\circ}$

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Jun 14, 2017

$\vec{r} = 13.5$ $\text{mi}$, at a bearing angle of ${50.4}^{\text{o}}$

#### Explanation:

We're asked to find the total displacement, both the magnitude and direction, of the ship after it leaves the port with the given conditions.

First, I'll explain what a bearing is.

A bearing is NOT a regular angle measure; normally, angles are measured anticlockwise from the positive $x$-axis, but bearing angles are measured clockwise from the positive $y$-axis.

Therefore, a bearing of ${34.0}^{\text{o}}$ indicates that this is an angle ${90.0}^{\text{o" - 34.0^"o" = color(red)(56.0^"o}}$ measured normally. We'll use this angle in our calculations.

We're given that the first displacement is $10.4$ $\text{mi}$ at an angle of ${56.0}^{\text{o}}$ (as calculated earlier). Let's split this up into its components:

${x}_{1} = 10.4 \cos {56.0}^{\text{o}} = 5.82$ $\text{m}$

${y}_{1} = 10.4 \sin {56.0}^{\text{o}} = 8.62$ $\text{m}$

Our second displacement is a simple $4.6$ $\text{mi}$ due east, that is, the positive $x$-direction. The components are thus

${x}_{2} = 4.6$ $\text{mi}$

${y}_{2} = 0$ $\text{mi}$

To find the total displacement from the port, we'll add these two vectors' components and use the distance formula:

$\Delta x = {x}_{1} + {x}_{2} = 5.82$ $\text{mi} + 4.6$ $\text{mi} = 10.42$ $\text{mi}$

$\Delta y = {y}_{1} + {y}_{2} = 8.62$ $\text{mi} + 0$ $\text{mi} = 8.62$ $\text{mi}$

$r = \sqrt{{\left({x}_{\text{total")^2 + (y_"total")^2) = sqrt((10.42"mi")^2 + (8.62"mi}}\right)}^{2}}$

= color(red)(13.5 color(red)("mi"

The direction of the displacement vector is given by

$\tan \theta = \frac{\Delta y}{\Delta x}$

so the angle is then

theta = arctan((Deltay)/(Deltax)) = arctan((8.62"mi")/(10.42"mi")) = 39.6^"o"

The question asked for the bearing angle, which is just this angle subtracted from ${90}^{\text{o}}$:

$\text{Bearing angle" = 90^"o" - 39.6^"o" = color(blue)(50.4^"o}$

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salamat Share
Jun 14, 2017

${8}^{\circ} 2 '$

#### Explanation:

Let say the distance of ship from port after travelled to the east $= x$
and the angle between a bearing of ${34}^{\circ}$ and to the east is $\left({90}^{\circ} - {34}^{\circ}\right) = {56}^{\circ}$

we use consine formula to find $x$

${x}^{2} = {4.6}^{2} + {10.4}^{2} - 2 \left(4.6\right) \left(10.4\right) \cos {56}^{\circ}$

${x}^{2} = 129.32 - 53.50 = 75.82$

$x = \sqrt{75.82} = 8.71$ mi

we use sinus formula to find the angle of displacement to east, let say
$= {y}^{\circ}$

$\frac{8.71}{\sin} {56}^{\circ} = \frac{4.6}{\sin} {y}^{\circ}$

$\sin {y}^{\circ} = \frac{4.6}{8.71} \cdot \sin {56}^{\circ}$

$\sin {y}^{\circ} = 0.4378$

$y = {25.97}^{\circ} = {25}^{\circ} 58 '$

therefore it bearing from the port $= \left({34}^{\circ} - {25.97}^{\circ}\right) = {8.03}^{\circ} = {8}^{\circ} 2 '$

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