Chapter 11, Problem 11.1P

To determine

The azimuth, bearing and length of the line connecting station shore and station rock.

Answer to Problem 11.1P

The azimuth of the line is $\text{ 20}°\text{50'2}\text{.14''}$

The bearing of the line is

   $\text{N 20}°\text{50'2}\text{.14'' E}$

The length of the line $164.8\text{ meters}$

Explanation of Solution

Given Information:

The coordinates of the stations in meters are:

Shore (379.241, 819.457)

Rock (437.854, 973.482)

The length of the line connecting the two station is

   $\begin{array}{l}L=\sqrt{{(X_{rock}-X_{shore})}^2+{(Y_{rock}-Y_{shore})}^2}\\ L=\sqrt{{(437.854-379.241)}^2+{(973.482-819.457)}^2}=164.8\text{ meters}\end{array}$

The azimuth of the line is

   $\begin{array}{l}{A}_Z={\tan }^{-1}(\frac{\Delta X}{\Delta Y})+C\\ A_Z={\tan }^{-1}(\frac{(437.854-379.241)}{(973.482-819.457)})+C\\ A_Z={\tan }^{-1}(\frac{58.61}{154.025})+0=0.\text{363178 = 20}°\text{50'2}\text{.14''}\end{array}$

The bearing of the line is

   $\text{N 20}°\text{50'2}\text{.14'' E}$

Conclusion:

The azimuth of the line is $\text{ 20}°\text{50'2}\text{.14''}$

The bearing of the line is

   $\text{N 20}°\text{50'2}\text{.14'' E}$

The length of the line $164.8\text{ meters}$