Chapter 8.2, Problem 37AYU

Solution

(a)

To find: The distance between each station from the ship.

The distance between each station from the ship are $135.57\text{miles}$ and $143.33\text{miles}$ .

Given information:

The Sea Coast Guard station able is located $150\text{miles}$ due south of Station Baker. The call to Station Able indicates that the ship is located $\text{N55}°\text{E}$ and the call to station Baker indicates that the ship is located $\text{S60}°\text{E}$ .

Explanation:

The distance between the station Baker and the ship is calculated as,

  $\begin{array}{l}\frac{b}{\sin 60°}=\frac{150}{\sin \left(180°-55°-60°\right)}\\ b=143.33\text{miles}\end{array}$

The distance between the station Able and the ship is calculated as,

  $\begin{array}{l}\frac{a}{\sin 55°}=\frac{150}{\sin \left(180°-55°-60°\right)}\\ a=135.57\text{miles}\end{array}$

Thus, the distance between each station from the ship are $135.57\text{miles}$ and $143.33\text{miles}$ .

(b)

To find: The time taken by the helicopter to reach the ship.

The time taken by the helicopter to reach the ship is $0.677\text{h}$ .

Given information:

The Sea Coast Guard station able is located $150\text{miles}$ due south of Station Baker. The call to Station Able indicates that the ship is located $\text{N55}°\text{E}$ and the call to station Baker indicates that the ship is located $\text{S60}°\text{E}$ . The speed of the helicopter is $200\text{miles}/\text{h}$ .

Explanation:

The distance between the station Able and the ship is nearest.

The time taken by the helicopter to reach the ship is calculated as,

  $\begin{array}{c}t=\frac{a}{v}\\ =\frac{135.57}{200}\\ =0.677\text{h}\end{array}$

Thus, the time taken by the helicopter to reach the ship is $0.677\text{h}$ .