Chapter 9.6, Problem 47PS

Solution

$\left(a\right)$

To find: To find how apart are the hikers after $1$ hour.

  $7\text{miles}$ .

Given information:

Starting at the same point in a forest, two hikers take different paths. The first hiker walks due north at a speed of $2$ miles per hour. The second hiker walks ${60}^{\circ }$ east of north at a speed of $3$ miles per hour.

Explanation:

The first hiker passes $2$ miles in one hour and the second hiker passes $3$

miles in one hour. It can be shown by the diagram below:

  

Use Law of Cosines to find the distance $d$ :

  $\begin{array}{l}d=2^2+3^2-2\times 2\times 3\times \cos \text{}{60}^{\circ }\\ d=4+9-2\times 2\times 3\times \frac{1}{2}\\ d=13-6=7\end{array}$

Hence, the required distance is $7\text{miles}\text{.}$

$\left(b\right)$

To find: The time when two hikers are out of range of each other.

  $3.78\text{hours}$

Given information:

The two hikers carry walkie-talkies with a range of $10$ miles. After how much time are the hikers out of range of each other.

Explanation:

Let after $t$ hours, the hikers are $10$ miles apart.

In $t$ hours, first hiker will be at the distance of $2t$ miles from a point $C$ .

In $t$ hours, second hiker will be at the distance of $3t$ miles from a point $C$ .

Let $c=2t,b=3t.$

By the law of Cosines, we have:

  $\begin{array}{c}{a}^2=b^2+c^2-2bc\cos A\\ {10}^2={\left(3t\right)}^2+{\left(2t\right)}^2-2\left(3t\right)\left(2t\right)\cos {60}^{\circ }\\ 100=9t^2+4t^2-2\left(3t\right)\left(2t\right)\frac{1}{2}\\ 100=13t^2-6t^2\\ 100=7t^2\\ t=\sqrt{\frac{100}{7}}\\ t=3\cdot 78\end{array}$

Hence, the two hikers are out of range of each other after $3.78\text{hours}$ .

$\left(c\right)$

To find: The time taken by the second hiker to meet the first hiker and to find in what direction should the second hiker should walk.

The second hiker will take $1.32\text{hr}$ approximately to meet the first hiker.

He should walk west of north at an angle of ${79.11}^{\circ }$ .

Given information:

Suppose after two hours the first hiker stops and tell the second hiker to meet her. How long will it take the second hiker to meet the first hiker $?$ In what direction should the second hiker walk $?$ Explain your reasoning.

Explanation:

In two hours, first hiker will be at the distance of $4$ miles from a point $C$ .

In two hours, second hiker will be at the distance of $6$ miles from a point $C$ .

By the law of Cosines, we have:

  $\begin{array}{c}{a}^2=b^2+c^2-2bc\cos A\\ a^2={\left(6\right)}^2+{\left(4\right)}^2-2\left(6\right)\left(4\right)\cos {60}^{\circ }\\ a^2=36+16-2\left(6\right)\left(4\right)\frac{1}{2}\\ a^2=52-24\\ a=\sqrt{28}\\ a=5\cdot 29\text{miles}\end{array}$

Since the second hikers walks at the speed of $3$ miles per hour, the time taken by him to cover the above distance will be:

  $\text{Time}=\frac{5.29}{4}\approx 1.32\text{hr}$

Hence, the second hiker will take $1.32\text{hr}$ approximately to meet the first hiker.

Now, find the direction in which the second hiker should walk by using the law of Cosine.

  $\begin{array}{c}{b}^2=a^2+c^2-2ac\cos B\\ 6^2={\left(\sqrt{28}\right)}^2+{\left(4\right)}^2-2\left(\sqrt{28}\right)\left(4\right)\cos B\\ 36=28+16-2\left(\sqrt{28}\right)\left(4\right)\cos B\\ \cos B=\frac{8}{2\left(\sqrt{28}\right)\left(4\right)}\\ \cos B=\text{0}\text{.1889}\\ \text{B=79}{\text{.11}}^{\circ }\end{array}$

Hence, the second hiker should walk west of north at an angle of ${79.11}^{\circ }$ .