# A Landowner with Triangular Piece of Flat Land| Vector Addition and Subtraction| Analytical Method| College Physics| Problem 3.20

### A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.61. She then correctly calculates the length and orientation of the third side C. What is her result? SOLUTION:

Since the figure is closed, we know that the sum of the three vectors is equal to 0. So, we have $\vec{A}+\vec{B}+\vec{C}=0$

Solving for C, we have $\vec{C}=-\vec{A}-\vec{B}$

Next, we solve for the x and y components of Vectors A and B.

For Vector A, the x and y components are $A_x=\left(80\:m\right)cos\left(21^{\circ} \right)=74.6864\:m$ $A_y=-\left(80\:m\right)sin\left(21^{\circ} \right)=-28.6694\:m$

For Vector B, the x and y components are $B_x=-\left(105\:m\right)sin\:11^{\circ} =-20.0349\:m$ $B_y=\left(105\:m\right)cos\:11^{\circ} =103.0709\:m$

Solve for the components of Vector C. $C_x=-A_x-B_x=-74.6864\:m\:-\left(-20.0349\:m\right)=-54.6515\:m$ $C_y=-A_y-B_y=-\left(-28.6694\:m\right)-\left(103.0709\:m\right)=-74.4015\:m$

Therefore, the length of Vector C is $C=\sqrt{\left(C_x\right)^2+\left(C_y\right)^2}$ $C=\sqrt{\left(-54.6515\:m\right)^2+\left(-74.4015\:m\right)^2}$ $C=92.3167\:m$

The direction of Vector C is $\theta =tan^{-1}\left(\frac{C_y}{C_x}\right)=tan^{-1}\left(\frac{74.4015\:m}{54.6515\:m}\right)=53.7^{\circ} \:\:South\:of\:West$