A smuggler's boat sets out on a course that makes a....

[btrfly24]

I just can't figure out where to start with this problem:

A smuggler's boat sets out at midnight at 20 mph on a course that makes a 40 degree angle with the shore. One hour later a DEA boat sets out also at 20 mph from a dock 80 miles up the coast to intercept the smuggler's boat. Find the angle for which the DEA boat will intercept the smuggler's boat and at what time will the interception occur?

This has me stumped. I'm leaning towards using the Law of Cosines somehow, but needing to account for the times and speeds is throwing me off. Could someone please help? Thanks so much in advance!

Catherine

 

[stapel]

A good place to start would probably be to draw a picture. For convenience sake, make the "shore" an horizontal line at the base of your drawing. Draw the forty-degree angle (it doesn't have to be "exact"), showing the smuggler's path, labelling the vertex of the angle as "S", for "smuggler".

The conditions of the exercise require that the smuggler be moving to the east (else the DEA agents could never catch him, since they're moving at the same rate). So label a point (call it "A", for "agents") off to the right of S, and label the segment SA as having length "80".

How far as the smuggler gone after one hour?

Starting at the time when the agents start out, so "t = 0" refers to when "a DEA boat sets out". Then what would be the expression for the distance covered by the agents in "t" hours? You have already found the (fixed) distance that the smuggler travelled in the first hour. What would be the expression for the total distance covered by the smuggler in "t" hours?

Draw a line from A representing the agents' path. Label the point where the two boats meet as "M", for "meeting". Use the distance expressions you just created to label the lengths of AM and SM.

You know the length of SA, and you know the measure of angle MAS. You have expressions for the lengths of AM and SM. Plug these into the Law of Cosines, and solve for "t".

Eliz.

 

[galactus]

The distance travelled by the smuggler is 20t. By the agent, 20(t-1).

Law of cosines:

\(\displaystyle \L\\(20t)^{2}+80^{2}-2(20t)(80)cos(40)=(20(t-1))^{2}\)

Solve for t. After you have that, find @(angle).

 

[soroban]

Hello, Catherine!

The trick is to get those speeds into the problem, right?

My approach is a variation of Galactus' . . .

A smuggler's boat sets out at midnight at 20 mph on a course
. . that makes a 40° angle with the shore.
One hour later a DEA boat sets out also at 20 mph from a dock 80 miles up the coast
. . to intercept the smuggler's boat.
Find the angle for which the DEA boat will intercept the smuggler's boat
. . and at what time will the interception occur?

                                      * I
                                  *  *
                         20t  *     *
                          *        *
                      *           *
                B o              * 20t
          20  *                 *
          *  40°            θ  *
      * - - - - - - - - - - - *
      A          80           D

At midnight, the smuggler starts at \(\displaystyle A\).
By 1 am, he is at point \(\displaystyle B:\;AB\,=\,20\) miles.
In the next \(\displaystyle t\) hours, he travels \(\displaystyle 20t\) miles to point \(\displaystyle I\).

At 1 am, the DEA starts at \(\displaystyle D\).
In the next \(\displaystyle t\) hours, it travels \(\displaystyle 20t\) miles to point \(\displaystyle I\).

Use the Law of Cosines to solve for \(\displaystyle t\).
. . \(\displaystyle (20t)^2\;=\;80^2\,+\,(20t\,+\,20)^2\,-\,2(80)(20t\,+\,20)\cos40^o\)

Once we know the three sides of the triangle,
. . we can determine \(\displaystyle \theta\:=\:\angle ADI\)

 

[TchrWill]

A smuggler's boat sets out at midnight at 20 mph on a course that makes a 40 degree angle with the shore. One hour later a DEA boat sets out also at 20 mph from a dock 80 miles up the coast to intercept the smuggler's boat. Find the angle for which the DEA boat will intercept the smuggler's boat and at what time will the interception occur?

................................................*D
............................................*..*
........................................*....*.*
...................................*.......*...*
...............................*.........*.....*
..........................*...........*.......*
.......................*.............*.........*
..................*................*...........*
...............*C...............*.............*
...........*...................*E..............*
......*.....................*............*.....*
*--------------------------------------*B
A
Assuming that the goal is to intercept the smuggler's boat as soon as possible:

AB is the shoreline = 80 miles
Angle DAB = 40º
AC = 20 miles at t = 1 hour
Draw, or envision, line CB
Bisect CB at E
Draw line ED
SInce both are traveling at 20 mph, CD and BD represent the distances traveled by both boats from t = 1 hour when the DEA boat departs.
Using AC == 20, AB = 80 and /_DAB = 40º, the Law of Cosines will derive BC being equal to 65.944 miles.
With equal speeds, CD = BD making triangle BCD isosceles.
Bisect BC at E.
CE = 65.944/2 = 32.972 miles
Angle ACD = arcsin(80(sin40)/65.944) = 128.759º making angle DCE = 51.241º.
CD = BD = 32.972/cos51.241º= 52.672 miles.
The time to intercept is then 52.672/20 = 2.633 hours.
The heading of the DEA boat derives from the sum of angles ABC and DBF which is 11.241 += 51.241 = 62.482º to the shore line toward the smuggler's position.

To recap, the smuggler travels 20 miles eastward at 40º to the shoreline in one hour from A to C
At t = 1 hour, the smuggler continues from C to D while the DEA boat travels westard, from a point 80 miles east of the smuggler, at a heading of 62.482º to the shoreline from B to D, where the DEA boat intercepts the smuggler at t = 3.633 hours from the smuggler's departure time.

Hope my arithmetic is right.

 

[btrfly24]

Thanks so much for everyone's help! It was nice to get more than one input on how to do this kind of problem. Catherine.