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## Class 11 Physics (India)

### Course: Class 11 Physics (India)>Unit 8

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:
$\begin{array}{rl}\mathrm{sin}\theta & =\frac{\text{opposite}}{\text{hypoteneuse}}\\ \\ \mathrm{cos}\theta & =\frac{\text{adjacent}}{\text{hypoteneuse}}\\ \\ \mathrm{tan}\theta & =\frac{\text{opposite}}{\text{adjacent}}\end{array}$