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:
Code:
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:
Code:
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