Hello, tishtash7!
Did you make a sketch?
A ship rounds a headland by sailing first four nautical miles on a course of 069°
then five nautical miles on a course of 295°.
Calculate the distance and bearing of its new position from its original position. Code:
C * Q
* * :
* * :
* * 65°:
* P * :
* : 46° * B
* : * :
* : * :
* :69°* 69°:
*: * :
* :
A R
The ship sails 4 miles from $\displaystyle A$ to $\displaystyle B.$
. . $\displaystyle AB \,=\,4,\;\angle PAB \,=\, 69^o \,=\,\angle ABR$
Then it sails 5 miles from $\displaystyle B$ to $\displaystyle C.$
. . $\displaystyle BC \,=\,5,\;\angle QBC \:=\:360^o - 295^o \:=\:65^o$
Hence: .$\displaystyle \angle CBA \:=\:180^o - 69^o - 65^o \:=\:46^o$
In $\displaystyle \Delta ABC$ use the Law of Cosinse:
. . $\displaystyle AC^2 \:=\:4^2 + 5^2 - 2(4)(5)\cos46^o \:=\:13.21366518$
Therefore:. $\displaystyle AC \:\approx\:\boxed{3.6\text{ miles}}$
Law of Cosines again: .$\displaystyle \cos(\angle CAB) \:=\:\frac{3.6^2 + 4^2-5^2}{2(3.6)(4)} \:=\:0.1375$
. . Hence: .$\displaystyle \angle CAB \:\approx\:82^o$
Then: .$\displaystyle \angle CAP \:=\:82^o - 69^o \:=\:13^o$
Therefore, the bearing is: .$\displaystyle 360^o - 13^o \:=\:\boxed{347^o} $