A sheep being chased by a sheep dog runs 25m north, then 40m west 30° south, and finally to a gate which is situated 20m north west (45°) from where the sheep started.
What was the distance and direction of the sheep's final run? And what total distance was the sheep chased?
Really having trouble about this problem, my classmate simply got out 2 pieces of graph paper and did the question to scale on the paper and found that the final run was approx. 22.5m E32°N . But that's kinda cheating isn't it? What's the formulaic and 'maths' way to approach this question?
Why would that be "cheating"? Were there specific instructions not to do that? A very similar method would be to set up a coordinate system, with the positive x-axis east, say, and the positive y-axis north, the origin at the point the sheep starts and meters as the unit length. The sheep first runs "25 m north" so he goes from (0, 0) to (0, 0+ 25)= (0, 25). It then runs "40 m west 30 s" so $(40 cos(210). 40 sin(210))= (-40 cos(30), -40 sin(30))= (-20\sqrt{3},-20)$ to $(-20\sqrt{3}, 5)$. Finally the sheep runs to a gate at $(-10\sqrt{2}, 10\sqrt{2})$. So the problem is to determine the length and direction of a vector from $(-20\sqrt{3}, 5)$ to $(-10\sqrt{2}, 10\sqrt{2})$. The length is easy- that is $\sqrt{(10\sqrt{2}- 20\sqrt{3})^2+ (5- 10\sqrt{2})^2}$. To determine the angle, subtract the point coordinates, $(10\sqrt{2}- 20\sqrt{3}, 5- 10\sqrt{2})$ and find $\theta$ such that $cos(\theta)= \frac{10\sqrt{2}- 20\sqrt{3}}{\sqrt{(10\sqrt{2}- 20\sqrt{3})^2+ (5- 10\sqrt{2})^2}}$ and $sin(\theta)= \frac{5- 10\sqrt{2}}{\sqrt{(10\sqrt{2}- 20\sqrt{3})^2+ (5- 10\sqrt{2})^2}}$