Chapter 14.5, Problem 35PPSE

Solution

a.

To calculate: The distance of the plane from city B if the pilot remains on this course for 20 mi. The pilot is flying from city A to city B which is 85 mi due north. After flying 20 mi, the pilot must change course and fly ${10}^{\circ }$ east of north to avoid a cloudbank.

The resultant answer distance is $25.541\text{ mi}$ .

Given information:

The pilot is flying from city A to city B which is 85 mi due north. After flying 20 mi, the pilot must change course and fly ${10}^{\circ }$ east of north to avoid a cloudbank.

Explanation:

First draw the diagram using the given information:

  

From the figure,

  $\begin{array}{c}B{C}^2={\left(20\sin \left({10}^{\circ }\right)\right)}^2+{\left(45-20\cos \left({10}^{\circ }\right)\right)}^2\\ B{C}^2={\left(20\sin \left({10}^{\circ }\right)\right)}^2+{45}^2-1800\cos \left({10}^{\circ }\right)+\left(20\cos \left({10}^{\circ }\right)\right)\\ B{C}^2={20}^2+{45}^2-1800\cos \left({10}^{\circ }\right)\\ B{C}^2=652.3\\ BC=25.541\text{ mi}\end{array}$

Therefore, the distance of the plane from city B was 25.541 mi.

b.

To state: The degrees that the pilot has to turn to the left to fly directly to city B and the number of degrees that are due north this course.

The pilot will have to turn ${92.185}^{\circ }$ to the left and the course is ${7.185}^{\circ }$ West to North.

Given information:

The pilot is flying from city A to city B which is 85 mi due north. After flying 20 mi, the pilot must change course and fly ${10}^{\circ }$ east of north to avoid a cloudbank.

Explanation:

The diagram is:

  

From the figure,

  $\begin{array}{c}\angle BCD={180}^{\circ }-{80}^{\circ }-{\sin }^{-1}\left(\frac{20\sin \left({10}^{\circ }\right)}{BC}\right)\\ ={180}^{\circ }-{80}^{\circ }-{\sin }^{-1}\left(\frac{20\sin \left({10}^{\circ }\right)}{25.541}\right)\\ ={180}^{\circ }-{80}^{\circ }-{\sin }^{-1}\left({7.815}^{\circ }\right)\\ ={92.185}^{\circ }\end{array}$

Therefore, the pilot will have to turn ${92.185}^{\circ }$ to left. This course is,

  ${\sin }^{-1}\left(\frac{20\sin \left({10}^{\circ }\right)}{25.541}\right)={7.185}^{\circ }$

Therefore, the course is ${7.185}^{\circ }$ West of North.