Answer:
[tex]B = 38[/tex]
Step-by-step explanation:
This question can be illustrated using the attachment and the required bearing will be calculated using cosine theorem;
[tex]b^2 = a^2 + c^2 - 2ac\ CosB[/tex]
In this case:
[tex]b = 5[/tex]
[tex]a = 7[/tex]
[tex]c = 8[/tex]
[tex]<B = ??[/tex]
Substitute these values in [tex]b^2 = a^2 + c^2 - 2ac\ CosB[/tex]
[tex]5^2 = 7^2 + 8^2 - 2 * 7 * 8\ CosB[/tex]
[tex]25 = 49 + 64 - 112\ CosB[/tex]
[tex]25 = 113- 112\ CosB[/tex]
Collect Like Terms
[tex]25 -113=- 112\ CosB[/tex]
[tex]-88=- 112\ CosB[/tex]
Divide through by -112
[tex]\frac{-88}{-112}= \frac{- 112\ CosB}{-112}[/tex]
[tex]\frac{-88}{-112}= CosB[/tex]
Reorder
[tex]Cos\ B = \frac{-88}{-112}[/tex]
[tex]Cos\ B = 0.7857[/tex]
Take arccos of both sides
[tex]B = cos^{-1}(0.7857)[/tex]
[tex]B = cos^{-1}(0.7857)[/tex]
[tex]B = 38[/tex] --- (approximated)
Hence, the bearing is approximately 38 degrees
