Question

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far is she from her camp base and what is her bearing from it

Answer 1

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Explanation:

The final position of the surveyor is represented by the following vectorial sum:

`\vec r = \vec r_(1) + \vec r_(2) + \vec r_(3)` (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

`(x,y) = r_(1)\cdot (\cos \theta_(1), \sin \theta_(1)) + r_(2)\cdot (\cos \theta_(2), \sin \theta_(2))` (1b)

Where:

`x, y` - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

`r_(1), r_(2)` - Length of each vector, in kilometers.

`\theta_(1), \theta_(2)` - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that
`r_(1) = 42\,km`,
`r_(2) = 28\,km`,
`\theta_(1) = 32^(\circ)` and
`\theta_(2) = 154^(\circ)`, then the resulting coordinates of the final position of the surveyor is:

`(x,y) = (42\,km)\cdot (\cos 32^(\circ), \sin 32^(\circ)) + (28\,km)\cdot (\cos 154^(\circ), \sin 154^(\circ))`

`(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]`

`(x,y) = (10.452, 34.531)\,[km]`

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

`\theta = \tan^(-1) (10.452\,km)/(34.531\,km)`

`\theta \approx 16.840^(\circ)`

And the distance from the camp is calculated by the Pythagorean Theorem:

`r = \sqrt{(10.452\,km)^(2)+(34.531\,km)^(2)}`

`r = 36.078\,km`

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).