# Find the vertical and horizontal component ? Steps please

Aug 30, 2017

${V}_{H} = 8.62$ and ${V}_{V} = 11.03$ to 2 dp

#### Explanation:

This is just trigonometry. We know that

$\sin \left(\theta\right) = \text{opposite"/"hypotenuse}$ and $\cos \left(\theta\right) = \text{adjacent"/"hypotenuse}$

So for all problems like this we can instantly see that the opposite corresponds to the vertical and the adjacent corresponds to the horizontal.

${V}_{H} = \left\mid \vec{V} \right\mid \cos \left(\theta\right)$ and ${V}_{V} = \left\mid \vec{V} \right\mid \sin \left(\theta\right)$

In this case, the magnitude is 14 and $\theta = 52.0$

${V}_{H} = 14 \cdot \cos \left(52\right) = 8.62$ and ${V}_{V} = 14 \cdot \sin \left(52\right) = 11.03$

Aug 30, 2017

${V}_{v} \approx 11.032$ (3 decimal places)
${V}_{h} \approx 8.619$ (3 decimal places)

#### Explanation:

You may recall SOHCAHTOA for a right angled triangle from trigonometry:

H = Hypotenuse, O = Opposite side, A = Adjacent side

$\sin \left(\theta\right) = \frac{O}{H} , \cos \left(\theta\right) = \frac{A}{H} , \tan \left(\theta\right) = \frac{O}{A}$

In this case, the horizontal component ${V}_{h}$ is adjacent to the angle, and the vertical component ${V}_{v}$ is opposite the angle. In both cases, $\theta$ is given as 52° and the hypotenuse is given as $14.0$.

We can note the two relationships to solve the vector into components:

$\sin \left(52\right) = \frac{O}{H} = \frac{{V}_{v}}{14}$

$\cos \left(52\right) = \frac{A}{H} = \frac{{V}_{h}}{14}$

rearranging for ${V}_{v}$ and ${V}_{h}$,

${V}_{v} = \sin \left(52\right) \cdot 14 \approx 11.03215055 \approx 11.032$ (3 decimal places)
${V}_{h} = \cos \left(52\right) \cdot 14 \approx 8.619260655 \approx 8.619$ (3 decimal places)