Bearing equation (closest approach to landmark along road)

[Horns36]

a family is traveling due west on a road that passes a famous landmark. At a given time the bearing to the landmark is N36*W and after the family travels 5 miles father the bearing is N 38*W. What is the closest the family will come to the landmark while on the road. Need help im totally confused

 

[tkhunny]

1) Draw a horizontal line. The "Road".
2) Label a direction "West".
3) Put a Monument "North" of the line.
4) To the "East" of the monument, on the "Road", put a car (a small one).
5) A little to the "West" of hte first car, but still "East" of the monument, put another car.
6) Farther "West", directly "South" of the monument, put a third car.
7) Draw a line segment between the monument and EACH car.
8) Ponder the three triangles - two of them Right Triangles.

 

[soroban]

Re: Bearing equation

Hello, Horns36!

A family is traveling due west on a road that passes a famous landmark.
At a given time the bearing to the landmark is N 36° W . . . . . . . . . These angles
and after the family travels 5 miles farther the bearing is N 38° W. . are reversed!
What is the closest the family will come to the landmark while on the road?

First, we have to understand the problem . . .

    L * 
      |    *
      |         *
      |              *
    d |                   *                   :
      |                        *              :
      |                             *         :
      |                             52°  * 38°:
    - + - - - - - - - - - - - - - - - - - - - * -
      C                                       A

At $\displaystyle A$, the bearing to the landmark $\displaystyle (L)$ is $\displaystyle N\,36^o\,W$.
Hence: $\displaystyle \,\angle LAC\,=\,52^o$

    L *
      | * 
      |   *           :
      |     *         :
    d |       *       :
      |         *     :
      |           *36°:
      |         54° * :
    - + - - - - - - - * - - - - - - - - - - - * -
      C               B          5            A

The car moves five miles to B and the bearing to L is $\displaystyle N\,36^o\,W$.
Hence: $\displaystyle \,\angle LBC\,=\,54^o.$


The diagram looks like this:

    L *
      | *  *
      |   *     *
      |     *        *
    d |       *           *
      |         *              *
      |           *                 * 
      |         54° *               52°  *
    - + - - - - - - - * - - - - - - - - - - - * -
      C       x       B          5            A

There are a number of ways to complete the problem . . . Here's one:

Let $\displaystyle x\,=\,CB$ and $\displaystyle d\,=\,LC.$

In right triangle $\displaystyle LCB:\;\tan54^o\,=\,\frac{d}{x}\;\;\Rightarrow\;\;d\,=\,x\cdot\tan54^o\;$ [1]

In right triangle \(\displaystyle LCA:\;\tan52^o\,=\,\frac{d}{x\,+\,5}\;\;\Rightarrow\;\;d\:=\x+5)\cdot\tan52^o\;\) [2]

Equate [1] and [2]: \(\displaystyle \:x\cdot\tan54^o\:=\x+5)\cdot\tan52^o\;\;\Rightarrow\;\;x\cdot\tan54^o\:=\:x\cdot\tan52^o\,+\,5\cdot\tan52^o\)

$\displaystyle \;\;$Then: $\displaystyle \:x\cdot\tan54^o\,-\,x\cdot\tan52^o\:=\:5\cdot\tan52^o\;\;\Rightarrow\;\;x(\tan54^o\,-\,\tan52^o)\:=\:5\cdot\tan52^o$

Hence: $\displaystyle \;x\:=\:\frac{5\cdot\tan52^o}{\tan544^o\,-\,\tan52^o}\:\approx\:66.36$

Substitute into [1]: $\displaystyle \:d\:=\:66.36\cdot\tan54^o \:\approx\:$91.34 miles.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This answer seemed to be way too big, but it checks out.

This means that, in the first diagram, we have a right triangle
$\displaystyle \;\;$ with $\displaystyle LC\,=\,91.336,\;\;CA\,=\,71.36$

Hence, the hypotenuse is: $\displaystyle \,LA\:=\:\sqrt{91.336^2\,+\,71.36^2}\:\approx\:116$


Imagine this scenario . . .

At point $\displaystyle A$, the driver gets out, takes a bearing to the landmark 116 miles away.
$\displaystyle \;\;$He must have a good compass and great binoculars!

He gets back into the car and drive another 5 miles to $\displaystyle B$
$\displaystyle \;\;$where he takes another bearing to the landmark 113 miles away.

He gets back into the car and does all this math.
He finds that the landmark is about 93.3 miles from the road.

Secure with this knowledge, he drives the remaining 66.36 miles to $\displaystyle C$
$\displaystyle \;\;$(taking another hour) and probably drives right by it.
(He never said that he intended to visit the landmark, did he?)

Then why did he go through all these calculations?
Obvious . . . he's a math major.