Answer:
The bearing needed to navigate from island B to island C is approximately 38.213º.
Step-by-step explanation:
The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:
(1)
Where:
- The distance from A to C, measured in miles.
- The distance from A to B, measured in miles.
- The distance from B to C, measured in miles.
- Bearing from island B to island C, measured in sexagesimal degrees.
Then, we clear the bearing angle within the equation:
![AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta](https://tex.z-dn.net/?f=AC%5E%7B2%7D-AB%5E%7B2%7D-BC%5E%7B2%7D%3D-2%5Ccdot%20AB%5Ccdot%20BC%5Ccdot%20%5Ccos%20%5Ctheta)
![\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}](https://tex.z-dn.net/?f=%5Ccos%20%5Ctheta%20%3D%20%5Cfrac%7BBC%5E%7B2%7D%2BAB%5E%7B2%7D-AC%5E%7B2%7D%7D%7B2%5Ccdot%20AB%5Ccdot%20BC%7D)
(2)
If we know that
,
,
, then the bearing from island B to island C:
![\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ccos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B%287%5Cmi%29%5E%7B2%7D%2B%288%5C%2Cmi%29%5E%7B2%7D-%285%5C%2Cmi%29%5E%7B2%7D%7D%7B2%5Ccdot%20%288%5C%2Cmi%29%5Ccdot%20%287%5C%2Cmi%29%7D%20%5Cright%5D)
![\theta \approx 38.213^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%5Capprox%2038.213%5E%7B%5Ccirc%7D)
The bearing needed to navigate from island B to island C is approximately 38.213º.