# A farmer wants to wall off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in the figure, where A = 4.99 km, B = 2.46 km, and C = 3.23 km and then correctly calculates the length and orientation of D?

## A farmer wants to wall off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in the figure, where A = 4.99 km, B = 2.46 km, and C = 3.23 km and then correctly calculates the length and orientation of the fourth side D. A: What is the length of the vector D in kilometers? B: What is the orientation of the vector D, in degrees W of S?

Nov 16, 2017

Magnitude: $D = \setminus \sqrt{{\left(- 1.22 k m\right)}^{2} + {\left(- 2.80 k m\right)}^{2}} = 3.05$ $k m$.
Orientation: $\setminus \theta = {180}^{o} + \arctan \left(\frac{- 2.80}{- 1.22}\right)$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = {180}^{o} + {66.4}^{o} = {246.4}^{o}$

#### Explanation:

Polar Representation: $\vec{V} = \left(V , \setminus \theta\right)$;
All angles must be measured counter-clockwise from the positive direction of the X-axis (usually East)

vec A = (4.99 km, 360^o-7.5^o) = (4.99 km, 352.5^o);
vec B = (2.46 km, 90^o + 16^o) = (2.46 km, 106^o);
vec C = (3.23 km, 180^o - 19^o = (3.23 km, 161^o);

Cartesian Representation: vec V = (V_x, V_y) = (V\cos\theta, V\sin\theta);
vec A = (4.99\times\cos352.5^o $k m , 4.99 \setminus \times \setminus \sin {352.5}^{o}$ km);
 \qquad = (4.95 $k m , - 0.65$ km);
vec B = (2.46\cos106^o $k m , 2.46 \setminus \sin {106}^{o}$ km);
\qquad = (-0.68 $k m , 2.4$ km);
vec C = (3.23\cos161^o $k m , 3.23 \setminus \sin {161}^{o}$ km);
\qquad = (-3.05 $k m , + 1.05$ km)

vec A + vec B + vec C = (1.22 $k m , 2.8$ km)

Zero Displacement Condition: $\vec{A} + \vec{B} + \vec{C} + \vec{D} = \vec{0}$
vec D = - (vec A + vec B + vec C) = (-1.22 $k m , - 2.80$ km)

Magnitude: $D = \setminus \sqrt{{\left(- 1.22 k m\right)}^{2} + {\left(- 2.80 k m\right)}^{2}} = 3.05$ $k m$.
Orientation: $\setminus \theta = {180}^{o} + \arctan \left(\frac{- 2.80}{- 1.22}\right)$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = {180}^{o} + {66.4}^{o} = {246.4}^{o}$

Note: In calculating the orientation, we recognize that the vector lies in the third quadrant because both the X and Y component are negative. Since the '$\setminus \arctan$' function only gives the angle between the vector and the nearest horizontal (negative X-axis), we have to explicitly add ${180}^{o}$ to make it stick to the convention of measuring all angles counterclockwise to the positive X-axis.