# Why is $u\cdot v/(||u||||v||)=\cos\theta$ not giving me the correct result? What am I doing wrong?

This might seem like a stupid question, but why is theta in the follow picture $$60^\circ$$, I understand how it is $$60^\circ$$ through simple trig, through $$2 \cos \theta$$, you can find $$\theta$$ to be $$60^\circ$$. Yet if you do the equation $$\frac{|u\cdot v|}{||u||\cdot ||v||} = \cos\theta$$ while $$v = [1,2]$$ and $$u = [1,0]$$, you get $$\theta \approx 63^\circ$$. What am I doing wrong? P.S. This comes from a question I got where:

Bert can swim at a rate of $$2$$ miles per hour in still water. The current in a river is ﬂowing at a rate of $$1$$ mile per hour. If Bert wants to swim across the river to a point directly opposite, at what angle to the bank of the river must he swim?

I know the answer is 60*, yet I am confused why it is 60 but not around 63. Thank you.

• I edited your post to improve the formatting. Please check that I haven't inadvertently changed the meaning. Commented Nov 28, 2020 at 15:54

Your vectors are wrong. In particular, the vector $$v$$ is not $$(1,2)$$ but $$(1,\sqrt{3})$$ so that its magnitude (i.e. the speed of swimming) would be $$2$$.
With $$v=(1,\sqrt{3})$$, you have $$u\cdot v=1$$ and $$|u|=1, |v|=2$$ so you get $$\cos\theta=1/2$$ as expected.
I see nothing wrong there. Indeed,$$\frac{u.v}{\|u\|.\|v\|}=\frac1{\sqrt5}\quad\text{and}\quad\arccos\left(\frac1{\sqrt5}\right)\simeq63.4^\circ.$$