Chapter 4.8, Problem 45E

a.

To determine

To find: The angle and bearing from $\text{A}$ to $\text{C}$ .

Answer to Problem 45E

The angle and bearing from $\text{A}$ to $\text{C}$ is$\angle \text{BCA}={54}^{\circ }$

Explanation of Solution

Given information:

Consider, the figure,

  

Calculation:

From the figure

Draw a vertical line $\text{AP}$

Then, ${\text{BAP = 32}}^{\text{o}}$

And ${\text{AAC = 90}}^{\text{o}}{\text{-32}}^{\text{o}}{\text{ = 58}}^{\text{o}}$

Also draw a vertical from $\text{C}$ to $\text{Q}$ .

So, that angle ${\text{QCB = 68}}^{\text{o}}$

Let, $\text{BCA}=\varphi$

Since, $\text{AP}$ and $\text{CQ}$ parallel

Then,

  $\begin{array}{l}\angle \text{PAC}+\angle \text{ACQ}={180}^{\circ }\\ \Rightarrow {58}^{\circ }+\varphi +{68}^{\circ }={180}^{\circ }\\ \Rightarrow \varphi ={54}^{\circ }\end{array}$

Hence, The angle and bearing from $\text{A}$ to $\text{C}$ is$\angle \text{BCA}={54}^{\circ }$ .

b.

To determine

To find: The angle and bearing from $\text{A}$ to $\text{C}$ .

Answer to Problem 45E

The angle and bearing from $\text{A}$ to $\text{C}$ is ${\text{N58}}^{\text{o}}\text{E}$ and $\text{68}\text{.82 m}\text{.}$

Explanation of Solution

Given information:

Consider, the figure,

  

Calculation:

From the figure

Draw a vertical line $\text{AP}$

Then, $\text{BAP}={32}^{\circ }$

And $\text{AAC = }{90}^{\circ }-{32}^{\circ }={58}^{\circ }$

Also draw a vertical from $\text{C}$ to $\text{Q}$ .

So, that angle $\text{QCB}={68}^{\circ }$

Let, $\text{BCA}=\varphi$

Bearing from $\text{A}$ to $\text{C}$ is ${\text{N58}}^{\text{o}}\text{E}$

And

  $\begin{array}{l}\tan {54}^{\circ }=\frac{\text{AB}}{50}\\ \Rightarrow \text{AB}=68.82\text{m}.\end{array}$

Hence, The angle and bearing from $\text{A}$ to $\text{C}$ is ${\text{N58}}^{\text{o}}\text{E}$ and $\text{68}\text{.82 m}\text{.}$