Chapter 11, Problem 11.1P

To determine

The length of line AB, azimuth of $AB$ , and bearing of AB.

Answer to Problem 11.1P

Thus, the length of $AB=367.539\text{m}$, the azimuth of $AB=31°18\text{'}24.36"$, and bearing of line AB is $\text{N}31°18\text{'}24.36"\text{E}$ .

Explanation of Solution

Given:

The coordinates of point A that is the station shore is $\left(246.873,659.457\right)$ and the coordinates of point B that is the station rock is $\left(437.854,973.482\right)$ .

Calculation:

Calculate the length of line $AB$ .

  $L_{\text{AB}}=\sqrt{{\left(X_{\text{B}}-X_{\text{A}}\right)}^2+{\left(Y_{\text{B}}-Y_{\text{A}}\right)}^2}$     ....... (I).

Here, $L_{\text{AB}}$ is the distance between point A and point B, $X_A$ , $X_B$ are the x-coordinates of point A and B respectively and $Y_{\text{A}}$ , $Y_{\text{B}}$ are the y-coordinates of point A and B respectively.

Substitute $246.873\text{m}$ for $X_{\text{A}}$ , $437.854\text{m}$ for $X_{\text{B}}$ , $659.457\text{m}$ for $Y_{\text{A}}$ and $973.482\text{m}$ for $Y_{\text{B}}$ equation (I).

  $\begin{array}{c}{L}_{\text{AB}}=\sqrt{{\left(437.854-246.873\right)}^2+{\left(973.482-659.457\right)}^2}\\ =\sqrt{{\left(190.981\right)}^2+{\left(314.025\right)}^2}\\ =\sqrt{36473.742+98611.701}\\ =367.539\text{m}\end{array}$

The length of line $AB=367.539\text{m}$ .

The azimuth of line AB is determined by using the formula,

${\text{Azimuth}}_{AB}={\tan }^{-1}\left(\frac{X-X_{\text{A}}}{Y_{\text{B}}-Y_{\text{A}}}\right)+C$     ....... (II)

Here, $Azimut{h}_{\text{AB}}$ is the azimuth of line AB, $X_A$ , $X_B$ are the x-coordinates of point A and B respectively, and $Y_{\text{A}}$ , $Y_{\text{B}}$ are the y-coordinates of point A and B respectively and $C$ is a constant.

Substitute $246.873\text{m}$ for $X_{\text{A}}$ , $437.854\text{m}$ for $X_{\text{B}}$ , $659.457\text{m}$ for $Y_{\text{A}}$ and $973.482\text{m}$ for $Y_{\text{B}}$ equation (II).

Since $\left(X-X_{\text{A}}\right)$ and $\left(Y_{\text{B}}-Y_{\text{A}}\right)$ are greater than zero, the value of $C=0°$ .

  $\begin{array}{c}{\text{Azimuth}}_{AB}={\tan }^{-1}\left(\frac{437.854-246.873}{973.482-659.457}\right)+0\\ ={\tan }^{-1}\left(\frac{190.981}{314.025}\right)\\ ={\tan }^{-1}\left(0.608\right)\\ =31°18\text{'}24.36"\end{array}$

Thus, the azimuth of $AB=31°18\text{'}24.36"$ .

The bearing of $AB$ is determined from the value of its azimuth.

Since $Azimut{h}_{\text{AB}}=31°18\text{'}24.36"$ , it lies between North and East, thus the bearing of line $AB$ is equal to $\text{N}31°18\text{'}24.36"\text{E}$ .

Conclusion:

Thus, the length of $AB=367.539\text{m}$, the azimuth of $AB=31°18\text{'}24.36"$, and bearing of line AB is $\text{N}31°18\text{'}24.36"\text{E}$ .