Respuesta :
Answer:
[tex]31^{\circ}[/tex]
Step-by-step explanation:
We have been given that vector v has a direction of (5,3). We are asked to find direction angle for v to nearest degree.
We know that direction of vector with components [tex](a,b)[/tex] can be determined by [tex]\theta=\text{tan}^{-1}(\frac{b}{a})[/tex].
Upon substituting the components of given vector in above formula, we will get:
[tex]\theta=\text{tan}^{-1}(\frac{3}{5})[/tex]
[tex]\theta=30.9637565^{\circ}[/tex]
Upon rounding to nearest degree, we will get:
[tex]\theta\approx 31^{\circ}[/tex]
Therefore, the direction angle for vector v is 31 degrees.
The direction of the path of the vector is measured relative to the
horizontal line or axis.
The direction angle for, v is approximately 31°.
Reasons:
The given vector can be expressed as follows;
v = 5·i + 3·j
Therefore, the vector extends 5 units in the x-direction and 3 units in the y-
direction.
The tangent of the angle, θ, formed by the direction of the vector, which is
the angle of the direction of the vector, is given as follows;
[tex]tan (\theta) = \dfrac{\mathrm{Distance \ moved \ in \ the \ y-direction}}{\mathrm{Distance \ moved \ in \ the \ x-direction}}[/tex]
Which gives;
[tex]tan (\theta) = \dfrac{3}{5}[/tex]
T
[tex]\theta = \arctan \left(\dfrac{3}{5} \right) \approx 30.964 ^{\circ}[/tex]
The direction angle for the vector, v, to the nearest degree θ ≈ 31°
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