Answer:
The angle is 75.52 degrees
Step-by-step explanation:
If there is a vector with the form a i + b j, the cosine of the angle between the given vector and the positive direction of the x - axis is calculated as:
cos θ = [tex]\frac{a}{\sqrt{a^{2}+b^{2}}}[/tex]
So, given the vector [tex]i +\sqrt{15} j[/tex], the cosine of the angle between the vector and the x-axis is:
cos θ = [tex]\frac{1}{\sqrt{1^{2}+\sqrt{15}^{2}}}=\frac{1}{\sqrt{16} }=0.25[/tex]
Therefore, the angle θ between the given vector and the positive direction of the x-axis is:
θ = [tex]cos^{-1}(0.25)=75.52[/tex]
Answer: 76 degrees.
The given vector is [tex] i +\sqrt{15} j[/tex].
We know that if a vector is xi + yj, then it makes an angle [tex]\tan^{-1}\left(\frac{y}{x}\right)[/tex] with the positive direction of the x-axis.
Here, x = 1 and [tex]y=\sqrt{15}[/tex].
So the angle is: [tex]\tan^{-1}\left(\frac{\sqrt{15}}{1}\right)=\tan^{-1}\left(\sqrt{15}\right)=76[/tex]
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