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### Course: High school physics>Unit 3

Lesson 2: Analyzing vectors using trigonometry

# Analyzing vectors using trigonometry review

Review the skills for analyzing vectors, including how to find horizontal and vertical components of vectors.

## Analyzing vectors with trigonometry

To simplify calculations for two-dimensional motion, we analyze the movement in the vertical direction separately from the horizontal direction. Since displacement, velocity, and acceleration are vector quantities, we can analyze the horizontal and vertical components of each using some trigonometry.

### Finding horizontal and vertical components

We can find the horizontal component ${A}_{x}$ and vertical component ${A}_{y}$ of a vector using the following relationships for a right triangle (see Figure 1a). $A$ is the hypotenuse of the right triangle.
${A}_{x}=A\mathrm{cos}\theta$
${A}_{y}=A\mathrm{sin}\theta$

### Determining the magnitude of the resultant

When we know the horizontal and vertical components, we can find the magnitude of their sum using the Pythagorean theorem (Figure 2).
$|A|=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}$

### Finding vector direction

To find the angle $\theta$ of the vector from the horizontal axis, we can use the horizontal component ${A}_{x}$ and vertical component ${A}_{y}$ in the trigonometric identity:
$\mathrm{tan}\theta =|\frac{{A}_{y}}{{A}_{x}}|$
We take the inverse of the $\mathrm{tan}$ function to find the angle $\theta$:
$\theta ={\mathrm{tan}}^{-1}|\frac{{A}_{y}}{{A}_{x}}|$

## Common mistakes and misconceptions

Sometimes people forget when to use $\mathrm{sin}$ or $\mathrm{cos}$ for calculating vector components. When in doubt, draw a right triangle and remember:
$\begin{array}{rl}\mathrm{sin}\theta & =\frac{\text{opposite}}{\text{hypoteneuse}}\\ \\ \mathrm{cos}\theta & =\frac{\text{adjacent}}{\text{hypoteneuse}}\\ \\ \mathrm{tan}\theta & =\frac{\text{opposite}}{\text{adjacent}}\end{array}$

For deeper explanations, see our video visualizing vectors in 2 dimensions.
To check your understanding and work toward mastering these concepts, check out the exercise on adding and decomposing vectors using trigonometry.

## Want to join the conversation?

• Two vectors A and B of magnitudes A = 30 units and B = 60 units respectively
are inclined to each other at angle of 60 degrees. Find the resultant vector.
• First, draw the vectors on any piece of paper. One way to approach this problem is to draw one vector that has an angle of elevation of 0 degrees, which just means that's parallel to the x-axis, and draw the other vector with an angle of elevation of 60 degrees. Let's assume that vector A is horizontal, and vector B is elevated at an angle of 60 degrees. Just connect both vectors with one vector being elevated 60 degrees above the other. If you were to draw the missing side of the triangle, you would see it's not a right triangle and therefore we can't use our basic SOH CAH TOA identities. Fortunately, we can use our beloved Law of Cosines because we know the length of two of the sides and the angle between them. I assume you know how to use it and therefore I will not explain it here, however, if this sounds unfamiliar to you, here is the URL link to the playlist of the Law of Sines and the Law of Cosines https://www.khanacademy.org/math/trigonometry/trig-with-general-triangles
Nevertheless, if we were to use the Law of Cosines to solve for the magnitude of the resultant, the answer should be (30)*(3^1/3) or 30 times the square root of 3.
;)
• should I have studied trigonometry before starting this course? Because I didn't, and I'm kind of confused now.
• Definitly. Trig is an integral part of this course.
• Why kicking a ball with 45 degree will travel further horizontally than 60 degree
• the horizontal distance is called the range. and R=(u^2sin(2thetha))/g. and since the largest value of sine is 1. and that only happens if thetha is 45.that means 2thetha would be 90 and the sine of 90 gives u one
• how do I know when to use the inverse of tan?
• When you have both sides ( opposite and adjacent) but no angle.
Normally you’re looking for a side or both given an angle. So you’d write it out as Sin 45 = opposite/3 (opposite/hypothenuse). But when you’re give two sides and looking for and angle you’d write it out — > tan 0= 4/3(opposite/adjacent). To solve that you’d write the inverse of tan (tan-1)
Which is tan-1(1.33) *divide before using inverse tan or else you’ll get a different answer.
You know to use the inverse of tan when given two sides and need to find the angle.
• why θ=tan^−1 ?I didn't catch :c
• Nice question!
First off -1 is not the exponent of tan there. tan is a function, a lot of functions have inverses. tan^-1 denotes the tan inverse function or the arctan function.
E.g. If we have tan(y)=x
Now if we apply the tan inverse function on both sides,
tan^-1(tan(y))=tan^-1(x) This gives:

y=tan^-1(x)

Hope that helps!
And that is a really good question, and its always nice to ask questions!
If anything was unclear or needs more explaining feel free to ask a follow up!
Cheers!
• = tan-1[Ay\Ax] what will further answer ?
• The horizontal distance is called the range.
(1 vote)
• why do you take modulus of ay/ax while finding theta
• i think it is really a good question. Actually it is just for sake of simplicity so as to not confuse those who are not aware of complex trigonometry. Because if ay/ax gets negative then theta comes to the angle where tangent is negative i.e.-2nd and 4th quadrant.
(1 vote)
• How do you tell if the horizontal or vertical components are negative/positive?