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?

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 ${90}^{\text{o}}$:

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

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:

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

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 $= \left({34}^{\circ} - {25.97}^{\circ}\right) = {8.03}^{\circ} = {8}^{\circ} 2 '$