# Thread: Calculating the acute angle between OX and the tangent to a curve.

1. ## Calculating the acute angle between OX and the tangent to a curve.

I don't know where to begin with this question. It's in the revision section in my textbook and I've had a look back through the book and I can't find anything helpful. I don't even know what OX is supposed to mean. Here it is:

Calculate, to the nearest degree, the acute angle between OX and the tangent to the curve y = {x}^{4} - {2x}^{3} + {3x}^{2} - 4x + 1 at the point where
x = 2.

According to the back of the book the answer is 86 degrees, but I would like to know how to get there.

2. Originally Posted by MAX1592
I don't even know what OX is supposed to mean. Here it is: Calculate, to the nearest degree, the acute angle between OX and the tangent to the curve y = {x}^{4} - {2x}^{3} + {3x}^{2} - 4x + 1 at the point where x = 2. According to the back of the book the answer is 86 degrees, but I would like to know how to get there.
In this case OX stands for the x-axis.
The slope of a line is the tangent of the angle the line makes with the x-axis in the positive direction.
So you need to find $\arctan\left(y^{\,\prime}(2)\right)$.
BTW: You have written the question wrong.
$y=x^4-2x^3+3x^2-4x+1$.
The answer is about $86.424^o$