Question Video: Finding the Horizontal Component of a Vector | Nagwa Question Video: Finding the Horizontal Component of a Vector | Nagwa

# Question Video: Finding the Horizontal Component of a Vector Physics • First Year of Secondary School

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The diagram shows a vector 𝐀 that has a magnitude of 22. The angle between the vector and the 𝑥-axis is 36°. Work out the horizontal component of the vector. Give your answer to 2 significant figures.

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### Video Transcript

The diagram shows a vector 𝐀 that has a magnitude of 22. The angle between the vector and the 𝑥-axis is 36 degrees. Work out the horizontal component of the vector. Give your answer to two significant figures.

Alright, we see this vector 𝐀 sketched out. And we’re told it has a length or magnitude of 22. Along with this, we know that the vector forms an angle of 36 degrees with the positive 𝑥-axis. Our goal is to solve for its horizontal component. This is equal to the horizontal projection of our vector onto this axis.

As we go about solving for the length of this orange line, let’s note that our dashed line intersects our horizontal axis at a right angle. In other words, we have here a right triangle. Here’s the hypotenuse, here’s another side, and here’s the third. In solving then for the horizontal component of our vector, we’re solving for one of the sides of this right triangle. We can do this using trigonometry.

Let’s remember that, given a right triangle, if we know one of the other interior angles, then we can define the sides of this right triangle as hypotenuse ℎ, the side opposite our angle 𝜃 𝑜, and the side adjacent to that angle 𝑎. Set up this way, it’s the adjacent side, what we’ve called 𝑎 over here, that we want to solve for to figure out our horizontal component.

Now, if we were to take the cos of this angle 𝜃, then that would equal the ratio of our adjacent side length to our hypotenuse. Or multiplying both sides of this equation by the hypotenuse, canceling that factor out on the right, we have that the adjacent side of our right triangle equals the cos of 𝜃 times ℎ. This relates to our situation with vector 𝐀 because in this case we know the length of our hypotenuse and we also know this angle. So we can actually say that the length of our triangle’s hypotenuse, 22, multiplied by the cos of our angle of 36 degrees is equal to what we’ll call 𝐴 sub 𝑥, the horizontal component of the vector 𝐀. When we enter this expression on our calculator and keep two significant figures, our answer is 18. This is the horizontal component of the vector 𝐀.

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