# Find the angle between the given vectors to the nearest tenth of a degree.

u = <6, 4>, v = <7, 5>

**Solution:**

If θ is the angle between two vectors u and v, then

cosθ = (u·v) / [lul lvl]

Given u = 6 i + 4 j

v = 7i + 5 j

Dot product of two vectors is u·v = a_{1}a_{2} + b_{1}b_{2}

u·v = (6)(7) + (4)(5) = 42 + 20

Therefore,

u.v = 62

llull = √[(a_{1})^{2 }+ (b_{1})^{2 }]

= √[(6)^{2 } + (4)^{2}]

=√36 + 16

lul = √36 + 16 = √52

lvl = √[(a_{2})^{2 } + (b_{2})^{2}]

= √[(7)^{2 } + (5)^{2}]

lvl = √49 + 25 = √74,

cos θ = 62 / [√52 √74]

⇒ cosθ = 62 / [√52 × √74]

⇒ cosθ = 62 / √3848

θ = cos^{-1}(0.999) = 0.9

Hence, nearest tenth will be 1 degree.

## Find the angle between the given vectors to the nearest tenth of a degree.

u = <6, 4>, v = <7, 5>

**Summary:**

The angle between the given vectors to the nearest tenth of a degree.u = <6, 4>, v = <7, 5> to the nearest tenth of a degree is 1 degree.

Math worksheets and

visual curriculum

visual curriculum