Question 1089471
you could solve this by creating right triangles, or you can solve using the following formula:


b^2 = a^2 + c^2 - 2 * a * c * cos(B)


this, i believe, is called the law of cosines.


here's a reference:


<a href = "http://www.mathwarehouse.com/trigonometry/law-of-cosines-formula-examples.php" target = "_blank">http://www.mathwarehouse.com/trigonometry/law-of-cosines-formula-examples.php</a>


here's a picture of my worksheet.


<img src = "http://theo.x10hosting.com/2017/080201.jpg" alt="$$$" </>


the triangle is labeled ABC.


A is the starting point.
B is the point 5 km due north.
C is the point 8 km due 30 degrees east of north.


angle B in the triangle is 150 degrees, since it is supplementary to the 30 degree angle.


side a is opposite angle A of the triangle.
side b is opposite angle B of the triangle.
sice c is opposite angle C of the triangle.


this fits the law of cosines since you are looking for the side opposite the known angle and you have the measure of the side adjacent to that angle.


the worksheet has the details of the calculations.


cos(150) is equal to -.8660254038


the rest is just solving the equation.


any questions, give me a shout.


otherwise, i believe you have all that you need.


another law that is sometimes useful is the law of sines.


a reference to that can be found here:


<a href = "http://www.mathwarehouse.com/trigonometry/law-of-sines/formula-and-practice-problems.php" target = "_blank">http://www.mathwarehouse.com/trigonometry/law-of-sines/formula-and-practice-problems.php</a>