# Thread: what direction to walk to make slope 26.6 deg

1. ## what direction to walk to make slope 26.6 deg

there is a 45 degree hill, z is up, y is north (the hill is sloped towards north), and x is east (hill is parallel to east). What direction should a hiker move so that the slope is 1/2 (i.e. 26.6 degrees)?

Intuitively I'd say the vector <1,1> because that's in between 45 degrees and zero

2. Originally Posted by superdude
there is a 45 degree hill, z is up, y is north (the hill is sloped towards north), and x is east (hill is parallel to east). What direction should a hiker move so that the slope is 1/2 (i.e. 26.6 degrees)?

Intuitively I'd say the vector <1,1> because that's in between 45 degrees and zero
If you walked up the hill (from the bottom to the top) you would be a distance above the bottom. Since the hill slopes at a 45 degrees angle you are also that horizontal distance from the bottom. (Its a 1:1 slope)
Call that vertical & horizontal distance 1.

If you walk along the top of the hill (parallel with the bottom) for a distance x.
The HORIZONTAL distance back to the starting point would be $\sqrt {1^2 + x^2}$.

You are looking for the horizontal distance, in this case, to be 2.

You have a triangle. Base (adjacent side) of length 1, and the hypotensue of length 2.

The angle you require (from the north direction straight up the hill):

$ArcCosine \left( \dfrac{1}{2} \right )$

3. Originally Posted by aidan

If you walk along the top of the hill (parallel with the bottom) for a distance x.
The HORIZONTAL distance back to the starting point would be $\sqrt {1^2 + x^2}$.

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If I walk x distance along top of hill, is that the same as moving x horizontal distance?

Wait a minute I think I see what's going on here