What is the bearing of A from C and what is the distance from B to C

Question

A girl starts from a point A and walks 285m to B on a bearing of 078 degrees.  She then walks due south to a point C which is 307m from A. What is the bearing of A from C and what is the distance from B to C.

Answer 1

There are at least two ways to solve this triangle problem. 

One method (used here) is  trigonometric / classical geometric

 

Bearings are measured from the North direction.   Thus the direction from A to B is

rotated 78 degrees away from North.  The direction from B to C is directly South. 

Thus the path from BC is parallel (actually anti-parallel) to the North direction.

Because of this parallel condition, the angle ABC is equal to 78 degrees by

alternate interior angles are equal. The law of cosines can be invoked to get

 307² = 285² + x² - 2 x 285 cos(78)

    where x is the distance BC.

 

This equation is quadratic in the unknown x.   The quadratic formula can be used

to determine x.   One value for x is negative (must be rejected)

the positive value from the quadratic formula is  x = 187.84

 

The angle, θ , the bearing from A to C  can be computed from the

law of sines.

 

    sin (θ) =(187.84 / 307) sin(78) =   0.5985 

   So θ =  36.76 degrees  (must be acute)

  this is angle BAC. 

 Since the sum of the interior angle of a triangle is 180,

the angle BCA =  65.24 degrees.  

This is the negative of the bearing of A from C.

So the bearing of A from C is  - 65.24