Chapter 4.8, Problem 27E

To determine

To find: How far apart are the ships.

Answer to Problem 27E

  $1^{st}$ ship is $=3071.93feet$ far away from the light home,

  $2^{nd}$ ship is $\left(3071.93+1933.3232\right)=5005.2332feet$ far away from the light home,

Explanation of Solution

Given information: An observer in a light home $350feet$ above the sea level observe two ships in the some vertical plane as the light home. The angles of deprerion to the ships are $4^{\circ }$ and ${6.5}^{\circ }$

Solution:

  

Here, the height of observe above sea level $=h=350feet$

  $\therefore$ Here, we find two triangle $△ABC,△ABD$

  $\therefore$ The angle of deprerion to ships are $4^{\circ }$ and ${6.5}^{\circ }$

  $\therefore$ The two angles are vertically opposite angles.

  $\therefore \angle ACB={6.5}^{\circ },\angle ADB=4^{\circ }$

  $\therefore$ We assume $1^{st}$ ship is $xfeet$ far apart from the light home

For triangle $△ABC\to$

  $\begin{array}{l}\therefore \tan \theta =\frac{perpendicular}{base}=\frac{h}{x}=\frac{350}{x}\\ \Rightarrow \tan {6.5}^{\circ }=\frac{350}{x}\\ \Rightarrow x=\frac{350}{\tan {6.5}^{\circ }}\\ \Rightarrow x=\frac{350}{0.113935}\\ \Rightarrow x=3071.9269\\ \Rightarrow x=3071.93feet\end{array}$

  $\therefore$ We assume $2^{nd}$ ship is $\left(x+y\right)feet$ far apart from the light home

For triangle $△ABD\to$

  $\begin{array}{l}\therefore \tan \theta =\frac{perpendicular}{base}=\frac{350}{x+y}\\ \Rightarrow \tan 4^{\circ }=\frac{350}{x+y}\\ \Rightarrow x+y=\frac{350}{\tan 4^{\circ }}\\ \Rightarrow \left(3071.93+y\right)=5005.2332\\ \Rightarrow y=1933.3032\end{array}$

Hence,

  $1^{st}$ ship is $=3071.93feet$ far away from the light home,

  $2^{nd}$ ship is $\left(3071.93+1933.3232\right)=5005.2332feet$ far away from the light home,