SOLUTION: A man travels 5 km due north from his starting point and then travels a further 8 km in a straight line in the direction 30 degrees east of north.How far is he from his starting po

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Question 1089471: A man travels 5 km due north from his starting point and then travels a further 8 km in a straight line in the direction 30 degrees east of north.How far is he from his starting position ?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
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:

http://www.mathwarehouse.com/trigonometry/law-of-cosines-formula-examples.php

here's a picture of my worksheet.

$$$

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:

http://www.mathwarehouse.com/trigonometry/law-of-sines/formula-and-practice-problems.php