Calculate the magnitude of the horizontal component of a vector that is 100 units long and oriented at 45 degrees.

Your question is somewhat twisted up. We would be justified in simply asking you to correct the question. But I think I know what you're talking about, so I'll try to answer it.
A horizontal vector is one that's oriented at an angle of zero degrees, not 45 degrees. I think what you want is the horizontal component of the vector that you're describing.
Its horizontal component is 100 cos(45) = 100 / √2 = 70.71 units .

Answered by AL2006 | 2024-06-10

To calculate the magnitude of the horizontal component of a vector that is 100 units long and oriented at a 45-degree angle, we use trigonometry. Specifically, we use the cosine function because it relates the adjacent side (the horizontal component) to the hypotenuse (the original vector) in a right-angled triangle.
Using the cosine function, we have:
Horizontal Component = Magnitude × Cosine(Angle)
So, if the vector's magnitude is 100 units and the angle is 45 degrees, we calculate the horizontal component as follows:
Horizontal Component = 100 × cos(45°)
We know that cos(45°) is √2/2. So the calculation becomes:
Horizontal Component = 100 × (√2/2)
This simplifies to 50√2, which is approximately 70.71 units.
Therefore, the magnitude of the horizontal vector is approximately 70.71.

Answered by JonathanAris | 2024-06-24

The magnitude of the horizontal component of a vector that is 100 units long and oriented at 45 degrees is approximately 70.71 units. This is calculated using the cosine function. The formula used is A x ​ = A ⋅ cos ( θ ) .
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Answered by AL2006 | 2024-10-13