Chapter 6.2, Problem 50E

To determine

Toapproximate:The distance by which the ships are apart at noon.

Explanation of Solution

Consider the figure shown below.

  

Let $A$ be the port location.

The position of ship at noon are $B$ and $C$ .

The distance travelled by one ship to $B$ is,

  $\begin{array}{c}c=12\left(3\right)\\ =36\end{array}$

The distance travelled by one ship to $C$ is,

  $\begin{array}{c}b=16\left(3\right)\\ =48\end{array}$

The included angle $A$ is calculated as,

  $\begin{array}{c}A=180°-53°-67°\\ =60°\end{array}$

Apply law of cosine to calculate the value of $a$ .

  $\begin{array}{c}{a}^2=b^2+c^2-2\left(b\right)\left(c\right)\sin A\\ a^2={48}^2+{36}^2-2\left(48\right)\left(36\right)\sin 60°\\ a^2=1872\\ a=43.27\end{array}$

Therefore, the distance between the ships is $43.27\text{miles}$ .