bearing problem: find distances, given bearings, etc.

[Guest]

The bearing of C from A is 226*. The bearing of B from A is 168*. The bearing of C from B is 316*. The distance from A to C is 290 km. Prove that the bearings form a right triangle among the 3 letters, and state any properties of angles used in the proof. Then determine the distance from A to B and the distance from B to C.

 

[galactus]

Here's a rough draft of your triangle. I used bearings off of due north(azimuth).

Note the interior angles of the triangle. What do they add up to?.

 

[Guest]

does anyone know how to determine the distances from A to B and from B to C?

 

[galactus]

Hello amandamandy:

I hope I didn't confuse you with the picture, I was trying to label everything.

Anyway, you have a right triangle, which makes things a little easier.

You know A to C is 290.

Therefore, 290/tan(32) is B to C

290/sin(32) is A to B

For that matter, once you find, say, B to C you could use Pythagoras to find the third side.

 

[soroban]

Hello, amandamandy!

The bearing of C from A is 226°.
The bearing of B from A is 168°.
The bearing of C from B is 316°.
The distance from A to C is 290 km.
Prove that the bearings form a right triangle among the 3 letters,
. . and state any properties of angles used in the proof.
Then determine the distance from A to B and the distance from B to C.

Here's my diagram:

                A
                *
              / :
        290 /46°:
          /     :
        /44°    :   N
    C * - - - - + - + - -
        \46°    S   :
          \         :
            \       :
              \     :
                \44o:
                  \ :
                    *
                    B

Since the bearing of AC is 226°, $\displaystyle \angle CAS\,=\,46^o$
$\displaystyle \;\;$In right triangle $\displaystyle ASC,\;\angle ACS\,=\,44^o$

Since the bearting of BC is 316°, $\displaystyle \angle CBN\,=\,44^o$
$\displaystyle \;\;$In right triangle $\displaystyle BNC,\;\angle BCN\,=\,46^o$

Therefore: $\displaystyle \,\angle ACB\:=\:44^o\,+\,46^o\:=\:90^o$
$\displaystyle \;\;\Delta ABC$ is a right triangle.


Rotating the right triangle, we have:

              B
              *
             /|
            / |
           /  |
          /   |
         /    |
        /     |
       /58°   |
      * - - - *
      A  290  C

Since $\displaystyle \,\tan58^o \,=\,\frac{BC}{290},\,$ we have: $\displaystyle \,BC\,=\,290\cdot\tan57^o\,\approx\,464.1$ km

Since $\displaystyle \,\cos57^o\,=\,\frac{290}{AB},\,$ we have: $\displaystyle \,AB\,=\,\frac{290}{\cos58^o}\,\approx\,547.3$ km

 

[galactus]

Cool, Soroban, we got the same thing.

I might add, as having been a surveyor, I tend to take azimuths off of

due north(the positive y-axis going clockwise) instead of the positive x

axis going counterclockwise, as it is mostly taught in math classes.


Nonetheless, the answer worked out the same.