Chapter 14.1, Problem 2P

To determine

To calculate: The magnitude to three significant digits and bearings to the nearest tenth of a degree and tofind the distance and bearing from the starting point.

A ship leaves port and sails $175$ km due southeastand then $168$ km due southwest,

Answer to Problem 2P

The distance is $\text{223 km}$ and the bearing point is ${48.9}^{\circ }$.

Explanation of Solution

Given information:

A ship leaves port and sails $175$ km due southeast and then $168$ km due southwest,

Formula used:

Pythagoras theorem of a right angled triangle,

  $H^2=B^2+P^2$

Where,

  $\begin{array}{l}H=hypotenues\\ B=base\\ P=perpendicular\end{array}$

Calculation:

Consider thatA ship leaves port and sails $175$ km due southeast and then $168$ km due southwest,

A line is $180$ degree so,

  $\begin{array}{c}\alpha =135-45\\ =90\end{array}$

Now by the low of cosines it is written as:

  $H^2=B^2+P^2-2BP\cos 90$

So it can be written as:

  $\begin{array}{c}{w}^2={175}^2+{168}^2-2\left(175\right)\left(168\right)cos90\\ w^2=30625+28224-58800\times 0\\ w=223\end{array}$

Now solve for $\sin$ .

  $\frac{\sin \theta }{168}=\frac{\sin 90}{223}$

Now apply the values of $\sin 90=1$ :

  $\frac{\sin \theta }{168}=\frac{1}{223}$

By cross multiplication:

  $\begin{array}{c}\frac{\sin \theta }{168}=\frac{1}{223}\\ \sin \theta =\frac{168}{223}\\ \sin \theta =0.753\\ \theta =48.9\end{array}$

Thus, the distance is $\text{223 km}$ and the bearing point is ${48.9}^{\circ }$ .