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?

3 Answers
Jun 14, 2017

vecr = 13.5 "mi", at a bearing angle of 50.4^"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^"o" indicates that this is an angle 90.0^"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 "mi" at an angle of 56.0^"o" (as calculated earlier). Let's split this up into its components:

x_1 = 10.4cos56.0^"o" = 5.82 "m"

y_1 = 10.4sin56.0^"o" = 8.62 "m"

Our second displacement is a simple 4.6 "mi" due east, that is, the positive x-direction. The components are thus

x_2 = 4.6 "mi"

y_2 = 0 "mi"

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

Deltax = x_1 + x_2 = 5.82 "mi" + 4.6 "mi" = 10.42 "mi"

Deltay = y_1 + y_2 = 8.62 "mi" + 0 "mi" = 8.62 "mi"

r = sqrt((x_"total")^2 + (y_"total")^2) = sqrt((10.42"mi")^2 + (8.62"mi")^2)

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

The direction of the displacement vector is given by

tantheta = (Deltay)/(Deltax)

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^"o":

"Bearing angle" = 90^"o" - 39.6^"o" = color(blue)(50.4^"o"

Jun 14, 2017

13.5"mi" at a bearing of 50.4^@

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^@ corresponds to a trigonometric angle of theta_1 =90^@-34^@ = 56^@

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

x = (10.4"mi")cos(56^@)
y = (10.4"mi")sin(56^@)

The trigonometric angle for the second heading is theta_2 = 90^@-90^@ = 0^@

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((10.4)^2 + (8.6)^2) ~~ 13.5"mi"

Its trigonometric angle is:

theta = tan^-1(y/x)

theta = tan^-1(8.6/10.4)

theta ~~ 39.6^@

The bearing angle is:

theta_b = 90^@-39.6^@ = 50.4^@

Jun 14, 2017

8^@ 2'

Explanation:

Let say the distance of ship from port after travelled to the east =x
and the angle between a bearing of 34^@ and to the east is (90^@ - 34^@) = 56^@

we use consine formula to find x

x^2 = 4.6^2 + 10.4^2 - 2(4.6)(10.4)cos 56^@

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^@

8.71/sin 56^@ = 4.6/sin y^@

sin y^@ = 4.6 /8.71 * sin 56^@

sin y^@ = 0.4378

y = 25.97^@ = 25^@ 58'

therefore it bearing from the port = (34^@ - 25.97^@) = 8.03^@ = 8^@ 2'