In the desert

A man wondering in the desert walks 5.7 miles in the direction S 26° W. He then turns 90° and walks 9 miles in the direction N 49° W. At that time, how far is he from his starting point, and what is his bearing from his starting point?

Final Answer:

m =  9.32 mi
φ =  274.8074 °

Step-by-step explanation:

$$\begin{gathered} a=5.7\text{mi} \\ b=9\text{mi} \\ \alpha =180+26=206{}^{\circ } \\ \beta =360-49=311{}^{\circ } \\ {x}_1=a\cdot \sin \alpha =a\cdot \sin 206°=5.7\cdot \sin 206°=5.7\cdot (-0.438371)=-2.49872\text{mi} \\ {y}_1=a\cdot \cos \alpha =a\cdot \cos 206°=5.7\cdot \cos 206°=5.7\cdot (-0.898794)=-5.12313\text{mi} \\ {x}_2=b\cdot \sin \beta =b\cdot \sin 311°=9\cdot \sin 311°=9\cdot (-0.75471)=-6.79239\text{mi} \\ {y}_2=b\cdot \cos \beta =b\cdot \cos 311°=9\cdot \cos 311°=9\cdot 0.656059=5.90453\text{mi} \\ x=x_1+x_2=(-2.4987)+(-6.7924)\doteq -9.2911\text{mi} \\ y=y_1+y_2=(-5.1231)+5.9045\doteq 0.7814\text{mi} \\ m=\sqrt{x^2+y^2}=\sqrt{(-9.2911{)}^2+0.7814^2}=9.32\text{mi} \end{gathered}$$

$$\begin{gathered} \tan \phi =y:x \\ x<0;y>0;\dots .IV.Quadrant \\ \Phi =\frac{180°}{\pi }\cdot \arctan (|y|/|x|)=\frac{180°}{\pi }\cdot \arctan (|0.7814|/|(-9.2911)|)\doteq 4.8074{}^{\circ } \\ \phi =270+\Phi =270+4.8074=274.8074^{\circ }=274°48^{\prime }27" \end{gathered}$$

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