Need Help with this Cal 3 question?

Let vector
a=i+2j
b=i
The dot product of a,b is
a*b=|a||b|cos(C), where C=the angle between
a & b
=>
(i+2j)*i=sqr(1+2^2)*1cos(C)
=>
cos(C)=1/sqr(5)
=>
C=63.435* (one possibility)
or
C=296.565* (another possibility)

Let θ be the angle between the vector i + 2 j and +x-axis.

√(1² + 2²) = √5

i + 2 j = (√5) [i (1/√5) + j (2/√5)]
i + 2 j = (√5) (i cosθ + j sinθ)

Hence, cosθ = 1/√5 > 0  and  sinθ = 2/√5 > 0
Hence, θ is in the first quadrant.

θ = cos⁻¹(1/√5)
The angle between the vector i + 2 j and +x-axis, θ = 64.43°

Strange that this would be a cal 3 question. 

You are given the two vectors (1,2) & (1,0), or take them as points, coordinates.  
Draw these on a grid, and the forming angle or triangle. 
Using highschool geometry, whats the angle ? Again, its just highschool geometry!

Then, try solving it via "Cal 3 math" by using the formula u•v = |u| |v| cos(θ).
You get the same answer? (you should). 

Done!

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them. 

Suppose we have 2 vectors a<a1,a2> and b<b1,b2> intersecting at an angle, t,say
Then cos(t) = adotb/[mod(a)*mod(b)], ...(1), where adotb = a1*b1 + a2*b2;
mod(a) = (a1^2 + a2^2)^(1/2), mod(b) = (b1^2 + b2^2)^(1/2).
Any vector running parallel to the x axis will have no vertical component. Such a
vector will be of form u<x,0>. So, in above problem we put a = <1,2>, b = <x,0>.
Then adotb = x, mod(a) = (1^2+2^2)^(1/2) = rt5, where rt ''is square root of''.;
mod(b) = x;
cos(t) = x/[rt5*x] = (1/rt5) = (1/5)rt5 andt = arccos[(1/5)rt5] = 63.43494882°.

the corresponding point is ( 1 , 2 ) so Θ = arctan 2