Find angle between the vector $\vec{a}=<1,2,2>$ and $\vec{b}=<3,4,0>$ in nearest degree
What I try:
$$\cos \theta=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{3+8}{3\times 5}=\frac{11}{25}$$
I did not understand how can I convert it into degrees.
When I use WolframAlpha it gives $\theta=1.152$. Please help me to find it in nearest degree angle.
We begin by finding the dot product of the two vectors. $$\vec{\mathbf{a}}\cdot\vec{\mathbf{b}}=(1)(3)+(2)(4)+(2)(0)=3+8+0=11$$ Next, let us find the magnitude of the first vector. $$\left\lVert\vec{\mathbf{a}}\right\rVert=\sqrt{(1)^2+(2)^2+(2)^2}=\sqrt{9}=3$$ Now, the magnitude of the second vector. $$\left\lVert\vec{\mathbf{b}}\right\rVert=\sqrt{(3)^2+(4)^2+(0)^2}=\sqrt{25}=5$$ We can now find the angle between these two vectors. $$\cos(\theta)=\frac{11}{3\cdot5}\implies\theta=\arccos\left(\frac{11}{15}\right)\approx42.83\text{°}$$ Thus, the angle measurement is approximately $43$ degrees.