# find the angle between the given vectors to the nearest tenth of degree, u =(4,2) and v =(5,4)

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### 2 Answers

The angle between two nonzero vectors is the angle `theta` ,` 0lt=thetalt=pi` , between their

respective standard position vectors. This angle can be found using the dot product:

`u . v = ||u|| ||v|| cos theta`

`rArr cos theta = (u . v)/(||u|| ||v||) ` -----------(i)

Here, `u =(4,2)` and `v =(5,4)` .

So, `u . v = (4)(5) + (2)(4) = 28`

`||u||=sqrt(4^2+2^2)=sqrt(20)=2sqrt5`

`||v||=sqrt(5^2+4^2)=sqrt(41)`

Substituting the values in equation (i) we get:

`cos theta=(28)/(2sqrt5*sqrt(41))`

`rArr cos theta=(28)/(2sqrt(205))`

**This implies that the angle between the two vectors is:**

`theta=arccos ((28)/(2sqrt(205)))~~12.1^o`

**Sources:**

The position vector `vec{ OU}=4veci+2vec j` .

The position vector `vec{OV}=5veci+4 vecj`

Let `theta ` be the angle between the vectors OU and OV.

`vec{OU}.vec{OV}=|vec{OU}||vec{OV}| cos(theta)`

But

`|vec {OU}|=sqrt(4^2+2^2)=2sqrt(5)`

`|vec{OV}|=sqrt(41)`

and

`vec{OU}.vec{OV}=5xx4+4xx2=28`

So,

`28=2sqrt(5) sqrt(41)cos(theta)`

`14/sqrt(205)=cos(theta)`

`theta=cos^(-1)((14sqrt(205))/205)`

`theta=12.09475708^o`