A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as$\mathit{A}$,$\mathit{B}$, and$\mathit{C}$in Figure 3.62, and then correctly calculates the length and orientation of the fourth side D. What is his result?

When two or more vectors with various magnitudes and directions are put together following the triangle law of vector addition, the resultant vector has the same impact as one vector.

• The magnitude of the vector$A$is, $A=4.70\text{km}$.
• The direction of the vector$A$ is $7.5°$south of east.
• The magnitude of the vector$B$is,$B=2.48\text{km}$.
• The direction of the vector$B$is $16°$west of north.
• The magnitude of the vector$C$is, $C=3.02\text{km}$.
• The direction of the vector $C$is $19°$north of west.

The horizontal component of the vector$A$is,

$A_x=A\cos \left(7.5°\right)$

Here$A$is the magnitude of the vector$A$.

Substitute $4.70\mathrm{km}$for $A$in the above expression, we get,

$\begin{array}{rcl}{A}_x& =& \left(4.70\text{km}\right)\times \cos \left(7.5°\right)\\ & =& 4.66\text{km}\\ & & \end{array}$

The horizontal component of the vector$B$is,

$B_x=-B\sin \left(16°\right)$

Here $B$is the magnitude of the vector$B$.

Substitute $2.48\mathrm{km}$for$B$ in the above expression, we get,

$\begin{array}{rcl}{B}_x& =& -\left(2.48\text{km}\right)\times \sin \left(16°\right)\\ & =& -0.68\text{km}\\ & & \end{array}$

The horizontal component of the vector$C$is,

$C_x=-C\cos \left(19°\right)$

Here $C$is the magnitude of the vector$C$.

Substitute $3.02\mathrm{km}$for $C$in the above expression, we get,

$\begin{array}{rcl}{C}_x& =& -\left(3.02\text{km}\right)\times \cos \left(19°\right)\\ & =& -2.86\text{km}\\ & & \end{array}$

The horizontal component of the resultant vector$D$ is,

$D_x=A_x+B_x+C_x$

Substitute $4.66\text{km}$for$A_x$, $-0.68\text{km}$for $B_x$, and$-2.86\text{km}$ for$C_x$ in the above expression, we get,

$\begin{array}{rcl}{D}_x& =& \left(4.66\text{km}\right)+\left(-0.68\text{km}\right)+\left(-2.86\text{km}\right)\\ & =& 1.12\text{km}\\ & & \end{array}$

The vertical component of the vector $A$is,

$A_y=-A\sin \left(7.5°\right)$

Here$A$is the magnitude of the vector$A$.

Substitute $4.70\mathrm{km}$for $A$in the above expression, we get,

$\begin{array}{rcl}{A}_y& =& -\left(4.70\text{km}\right)\times \sin \left(7.5°\right)\\ & =& -0.61\text{km}\\ & & \end{array}$

The vertical component of the vector$B$ is,

$B_y=B\cos \left(16°\right)$

Here $B$is the magnitude of the vector$B$.

Substitute $2.48\mathrm{km}$for $B$in the above expression, we get,

$\begin{array}{rcl}{B}_y& =& \left(2.48\text{km}\right)\times \cos \left(16°\right)\\ & =& 2.38\text{km}\\ & & \end{array}$

The vertical component of the vector$C$ is,

$C_y=C\sin \left(19°\right)$

Here $C$is the magnitude of the vector$C$ .

Substitute$3.02\mathrm{km}$ for $C$in the above expression, we get,

$\begin{array}{rcl}{C}_y& =& \left(3.02\text{km}\right)\times \sin \left(19°\right)\\ & =& 0.98\text{km}\\ & & \end{array}$

The vertical component of the resultant vector $D$is,

$D_y=A_y+B_y+C_y$

Substitute $-0.61\text{km}$for$A_y$ , $2.38\text{km}$for $B_y$, and$0.98\text{km}$ for $C_y$in the above expression, we get,

$\begin{array}{rcl}{D}_y& =& \left(-0.61\text{km}\right)+\left(2.38\text{km}\right)+\left(0.98\text{km}\right)\\ & =& 2.75\text{km}\\ & & \end{array}$

The magnitude of the resultant vector $D$is,

$D=\sqrt{D_x^2+D_y^2}$

Substitute$1.12\mathrm{km}$ for $D_x$, and $2.75\text{km}$for $D_y$in the above expression, we get,

$\begin{array}{rcl}D& =& \sqrt{{\left(1.12\text{km}\right)}^2+{\left(2.75\text{km}\right)}^2}\\ & =& 2.97\text{km}\\ & & \end{array}$

The direction of the resultant vector$D$ is,

$\theta ={\tan }^{-1}\left(\frac{D_x}{D_y}\right)$

Substitute for $1.12\text{km}$, $D_x$and for in the above expression, we get,

role="math" localid="1668687995126" $\theta ={\tan }^{-1}\left(\frac{1.12}{2.75}\right)$

$\theta =2216°$

Hence, the length of the fourth side$D$ is$2.97\mathrm{km}$ and oriented to $22.16°$the west of south.