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?



Since the figure is closed, we know that the sum of the three vectors is equal to 0. So, we have 


Solving for C, we have


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. 



Therefore, the length of Vector C is




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