Chapter 14.1, Problem 1P

To determine

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

A plane flies $285$ km due west and $320$ km due north.

Answer to Problem 1P

The distance is $429\text{ km}$ and the bearing point is ${48.2}^{\circ }$

Explanation of Solution

Given information:

A plane files $285$ km due west and $320$ km due north.

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 the plane files $285$ km due west and $320$ km due north.

Recall that the Pythagoras theorem,

  $H^2=B^2+P^2$

Here the base and perpendicular and the base are: $285\text{ and 320, }x$ .

Apply the values in the formula as:

  $\begin{array}{c}{H}^2=B^2+P^2\\ x^2={285}^2+{320}^2\\ x^2=81225+102400\\ x^2=183625\end{array}$

Then the value of $x$ is:

  $\begin{array}{c}{x}^2=183625\\ x=428.51\\ x\approx 429\end{array}$

So the hypotenuse is $429$ or can say that the distance between its points is $429$ km.

Now to find the bearing from the starting point by the help of sine law that is:

  $\frac{\sin \theta }{P}=\frac{\sin {90}^{\circ }}{H}$

Now apply the values in the formula with the value of $\sin 90=1$ :

  $\begin{array}{c}\frac{\sin \theta }{P}=\frac{\sin {90}^{\circ }}{H}\\ \frac{\sin \theta }{320}=\frac{\sin 90}{429}\\ \frac{\sin \theta }{320}=\frac{1}{429}\end{array}$

By cross multiplication:

  $\begin{array}{c}\frac{\sin \theta }{320}=\frac{1}{429}\\ \sin \theta =\frac{320}{429}\\ \sin \theta =0.746\\ \theta =48.2\end{array}$

Thus, the distance is $429\text{ km}$ and the bearing point is ${48.2}^{\circ }$ .