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
Nathan L.
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"#

Douglas K.
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^@#

salamat
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'#

Related questions
Impact of this question
21930 views around the world
You can reuse this answer
Creative Commons License