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Answer:
The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).
Step-by-step explanation:
The final position of the surveyor is represented by the following vectorial sum:
[tex]\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}[/tex] (1)
And this formula is expanded by definition of vectors in rectangular and polar form:
[tex](x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})[/tex] (1b)
Where:
[tex]x, y[/tex] - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.
[tex]r_{1}, r_{2}[/tex] - Length of each vector, in kilometers.
[tex]\theta_{1}, \theta_{2}[/tex] - Bearing of each vector in standard position, in sexagesimal degrees.
If we know that [tex]r_{1} = 42\,km[/tex], [tex]r_{2} = 28\,km[/tex], [tex]\theta_{1} = 32^{\circ}[/tex] and [tex]\theta_{2} = 154^{\circ}[/tex], then the resulting coordinates of the final position of the surveyor is:
[tex](x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})[/tex]
[tex](x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km][/tex]
[tex](x,y) = (10.452, 34.531)\,[km][/tex]
According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:
[tex]\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}[/tex]
[tex]\theta \approx 16.840^{\circ}[/tex]
And the distance from the camp is calculated by the Pythagorean Theorem:
[tex]r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}[/tex]
[tex]r = 36.078\,km[/tex]
The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).