Question

how can we construct 17 degree angle​

Answer 1

Answer:

Yes

Step-by-step explanation:

Suppose that a  17∘  angle could be constructed using compass and straight edge. Since a  60∘  angle is constructible and any angle can be bisected using only these tools, a  15∘  angle is constructible (by bisecting a  60∘  angle twice). By embedding this in our  17∘  angle, we construct an angle of  2∘  and by reproducing ten adjacent copies of a  2∘  angle, we obtain a  20∘  angle. Finally, by taking a point on one ray of this angle unit distance from the vertex and dropping a perpendicular to the other ray, we construct a line segment of length  cos20∘ .

Now we apply some field theory (as found in most current introductory abstract algebra texts). Let  α=cos20∘ . Taking  θ=20∘ , the identity  cos3θ=4cos3θ−3cosθ  implies that  α  is a root of the polynomial  8x3−6x−1 . This polynomial has no rational root and so, because it is of degree 3, it is irreducible over  Q . Therefore, the extension field  Q(α)  has degree 3 over  Q . On the other hand, because  α  is constructible, it lies in an extension  K  of  Q  of degree  2n  for some  n . Because  |Q(α):Q|  divides  |K:Q| . we have a contradiction and this proves that, in fact, an angle of  17∘  is not constructible with compass and straight edge.