Chapter 4.8, Problem 75E

To determine

To find:The two bearings that are perpendicular to ${\text{N32}}^{\text{o}}\text{E}$ .

Answer to Problem 75E

The two bearings perpendicular to ${\text{N32}}^{\text{o}}\text{E}$ are ${\text{N58}}^{\text{o}}\text{W}$ and ${\text{N122}}^{\text{o}}\text{S}$ .

Explanation of Solution

Given information:The direction to which the two bearings are to be found is ${\text{N32}}^{\text{o}}\text{E}$ .

  ${\text{N32}}^{\text{o}}\text{E}\text{means}\text{an}\text{angle}\text{of}{\text{32}}^{\circ }\text{from}\text{North}\text{towards}\text{East}\text{.}$

Since the two bearings are perpendicular to ${\text{N32}}^{\text{o}}\text{E}$ , therefore they will make an angle of ${90}^{\circ }\text{to}\text{this}\text{direction}\text{as}\text{shown}\text{in}\text{the}\text{sketch}\text{below}$

  

It can be observed in the figure that $\text{A}\text{is}{\text{N32}}^{\text{o}}\text{E}\text{and}\text{the}\text{two}\text{perpendiculars}\text{to}\text{OA}\text{are}\text{point}\text{B}\text{and}\text{C}$ .

Therefore the bearing for point B is

  $\begin{array}{l}={90}^{\circ }-{32}^{\circ }\\ ={58}^{\circ },\text{towards}\text{west}\\ \text{that}\text{means}{\text{N58}}^{\text{o}}\text{W}\end{array}$

The bearing for point C is

  $\begin{array}{l}={90}^{\circ }+{32}^{\circ }\\ ={122}^{\circ },\text{towards}\text{south}\\ \text{that}\text{means}{\text{N122}}^{\text{o}}\text{S}\end{array}$

Hence the two bearings perpendicular to ${\text{N32}}^{\text{o}}\text{E}$ are ${\text{N58}}^{\text{o}}\text{W}$ and ${\text{N122}}^{\text{o}}\text{S}$ .