Question

An airplane, whose airspeed is 580 km/h, is supposed to fly in a straight path$\mathbf{38}\mathbf{.}\mathbf{0}\mathbf{°}$N of E. However, a steady 82 km/h wind is blowing from the north. In what direction should the plane head? [Hint: Use the law of sines, Appendix A-7.]

Short answer

The plane should head to $44.40°$north of east.

Step-by-step solution

Step 1. Given data

According to the law of sines,the ratio of each side of a plane triangle to the sine of the opposite angle is the same for all three sides and angles.You can use the law of sines to compute the remaining sides of a triangle when two angles and a side are known.

Given data:

The speed of the airplane relative to air is $v_{\mathrm{p}}=580\frac{\mathrm{km}}{\mathrm{h}}$.

The rate of the wind relative to the ground is $v_{\mathrm{w}}=82\frac{\mathrm{km}}{\mathrm{h}}$from the north.

The path of the plane is $38.0°$north of east.

Assumptions:

Assume that $v_{\mathrm{r}}$is the plane's speed relative to the ground, and the plane's angle is $\theta °$north of east.

Step 2. Representation of the velocity vectors

Now, the representation of the velocity vectors is shown below as follows:

Step 3. Use the law of sines

Now, using the law of sines in the triangle ABC,

$\begin{array}{c}\frac{v_{\mathrm{p}}}{\sin 128°}=\frac{v_{\mathrm{w}}}{\sin \varphi }\\ \sin \varphi =\frac{v_{\mathrm{w}}}{v_{\mathrm{p}}}\sin 128°\\ \varphi ={\sin }^{-1}\left(\frac{v_{\mathrm{w}}}{v_{\mathrm{p}}}\sin 128°\right)\end{array}$

Now, substituting the numerical values in the above equation,

$\begin{array}{c}\varphi ={\sin }^{-1}\left(\frac{82\frac{\mathrm{km}}{\mathrm{h}}}{580\frac{\mathrm{km}}{\mathrm{h}}}\sin 128°\right)\\ =6.40°\end{array}$

Now, the angle of the plane should be

$\begin{array}{c}\theta =38.0°+\varphi \\ =38.0°+6.40°\\ =44.40°\end{array}$

Hence, the plane should head to $44.40°$north of east.