A man 7

A man wandering in the desert walks 3.8 miles in the direction of S 44° W. He then turns and walks 2.2 miles toward N 55° W. At that time, how far is he from his starting point? (Round your answer to two decimal places.)

Final Answer:

d =  4.68 mi

Step-by-step explanation:

$$\begin{gathered} \alpha =180+44=224{}^{\circ } \\ a=3.8\text{mi} \\ \beta =360-55=305{}^{\circ } \\ b=2.2\text{mi} \\ {x}_1=a\cdot \sin \alpha =a\cdot \sin 224°=3.8\cdot \sin 224°=3.8\cdot (-0.694658)=-2.6397\text{mi} \\ {y}_1=a\cdot \cos \alpha =a\cdot \cos 224°=3.8\cdot \cos 224°=3.8\cdot (-0.71934)=-2.73349\text{mi} \\ {x}_2=b\cdot \sin \beta =b\cdot \sin 305°=2.2\cdot \sin 305°=2.2\cdot (-0.819152)=-1.80213\text{mi} \\ {y}_2=b\cdot \cos \beta =b\cdot \cos 305°=2.2\cdot \cos 305°=2.2\cdot 0.573576=1.26187\text{mi} \\ x=x_1+x_2=(-2.6397)+(-1.8021)\doteq -4.4418\text{mi} \\ y=y_1+y_2=(-2.7335)+1.2619\doteq -1.4716\text{mi} \\ d=\sqrt{x^2+y^2}=\sqrt{(-4.4418{)}^2+(-1.4716{)}^2}=4.68\text{mi} \end{gathered}$$

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