Question

In Fig.$\mathbf{3}\mathbf{-}\mathbf{27}$, a heavy piece of machinery is raised by sliding it a distance$\mathit{d}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{m}$along a plank oriented at angle$\mathit{\theta }\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$to the horizontal. How far is it moved (a) vertically and (b) horizontally?

Short answer

$\mathbf{a}\mathbf{)}$ Vertical height is$4.28\mathrm{m}$

$\mathbf{b}\mathbf{)}$Horizontal distance is$11.7\mathrm{m}$

Step-by-step solution

To understand trigonometric functions

Trigonometry is used to compute heights of structures, construct maps, and locate islands using longitudes and latitudes. In this problem,using the following trigonometric relation and the given incline length, the distance moved by the heavy piece vertically and horizontally can be computed.

$$\begin{gathered} h=d\sin \theta \\ x=d\cos \theta \\ \left(\mathrm{ii}\right) \end{gathered}$$

Where,$h$and$x$are the height and horizontal distance respectively.

Given

$\mathrm{d}=12.5\mathrm{m}\mathrm{and}\theta =20.0°$

To find vertical height h 

Using equation (i), $h$can be written as,

$$\begin{gathered} h=d\sin \theta =12.5\times \sin \left(20.0°\right) \\ h=4.28\mathrm{m} \end{gathered}$$

To find horizontal distancex

Using equation (ii), $x$can be written as,

$$\begin{gathered} x=d\cos \theta =12.5\times \cos \left(20.0°\right) \\ x=11.7\mathrm{m} \end{gathered}$$