Chapter 12.5, Problem 30PPS

To determine

To find: How many minutes did the entire trip take.

Answer to Problem 30PPS

Total time for the entire trip $=0.0246hrs$

Explanation of Solution

Given information:

  $\begin{array}{l}a=0.15\\ c=0.25\\ B={130}^0\end{array}$

Use the Law of cosines to find the value of missing side .

We get

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos B\\ \Rightarrow b^2={\left(0.15\right)}^2+{\left(0.25\right)}^2-2\left(0.15\right)\left(0.25\right)\cos \left({130}^0\right)\\ \Rightarrow b^2=0.0225+0.0625-0.075\left(-0.643\right)\\ \Rightarrow b^2=0.0225+0.0625+0.048\\ \Rightarrow b^2=0.133\\ \Rightarrow b=\sqrt{0.133}\\ \Rightarrow b=0.36\end{array}$

Time taken for the trip from point A to point B is $\frac{0.25}{28}=0.0089$ hours

Time taken for the trip from point B to point C is $\frac{0.15}{28}=0.0054$ hours

Time taken for the trip from point C to point A is $\frac{0.36}{35}=0.0103$ hours

Therefore,

Total time for the entire trip $=0.0089+0.0054+0.0103=0.0246hrs$