How far has the checker piece moved? A checker board has squares of sides 4.0 cm. A piece does a triple move such that it ends up two squares right and six squares forward from where it began. What is the total displacement and direction of the piece taking into account significant figures?
Using Pythagoras the answer would be: $$d = \sqrt{(2*4.0)^2 + (6*4.0)^2} = 25.2982...cm$$
Which when reduced to two significant figures gives $25cm$.
The direction would be: $$\theta=\tan^{-1}(\frac{6*4.0}{2*4.0})=71.5650...^o$$
Which when reduced to two significant figures would be $72^o$.
Is this correct?
We were given the distance 4.0 to two significant figures. That means at most this number could be 4.049999 recurring, and at least this number could be 3.95.
Finding the distance and angle for both the max-case scenario and the min-case scenario we get:
Max-case scenario: $$d=25.6144...cm ; \theta = 71.9905...^o$$
Min-case scenario: $$d=24.9819...cm ; \theta = 71.1310...^o$$
Question 1: The angle would then be $71.5607...\pm 0.4298... ^o$. Which would be $71.6\pm 0.4^o$ Is this correct? 
Question 2: How should I present the magnitude of displacement? Taking the entire uncertainty range it seems it should be: $25.2982... \pm 0.3165...cm$. So maybe $25.3\pm 0.3cm$? What is the correct scientific way to answer this?

Actually, the max case of the angle is
$$\tan^{-1}\left(\frac{6\cdot 4.05}{2\cdot 3.95}\right)$$
(and I'll let you figure the min case)
because you shouldn't assume that all sides of the squares are completely equal!

Other than that, you are correct. Your initial calculations are good, and your final result for the length is also correct: $25.3\pm 0.3$cm.
