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 \(80.0{\rm{ m}}\) long and the next to be \(105\;{\rm{m}}\). These sides are represented as displacement vectors \({\rm{A}}\) from \({\rm{B}}\) in Figure 3.61. She then correctly calculates the length and orientation of the third side \({\rm{C}}\). What is her result?

The breaking of a single vector into two or more vectors of differing magnitude and direction that, when combined, provide the same effect as a single vector generates is known as vector resolution.

• The magnitude of the vector , \(A = 80\;{\rm{m}}\).
• Angle made by the vector \({\rm{A}}\) with the horizontal axis is \({21^ \circ }\).
• The magnitude of the vector \({\rm{B}}\), \(B = 105\;{\rm{m}}\).
• Angle made by the vector \({\rm{B}}\) with vertical axis is \({11^ \circ }\).

The horizontal component (\(x\) component) of the vector \({\rm{A}}\) is,

\({A_x} = A\cos \left( {{{21}^ \circ }} \right)\)

Here \(A\) is the magnitude of the vector \({\rm{A}}\).

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{A_x} &= \left( {80\;{\rm{m}}} \right) \times \cos \left( {{{21}^ \circ }} \right)\\ &= 74.69\;{\rm{m}}\end{aligned}\)

The horizontal component (\(x\) component) of the vector \({\rm{B}}\) is,

\({B_x} = - B\sin \left( {{{11}^ \circ }} \right)\)

Here \(B\) is the magnitude of the vector \({\rm{B}}\).

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{B_x} &= - \left( {105\;{\rm{m}}} \right) \times \sin \left( {{{11}^ \circ }} \right)\\ &= - 20.03\;{\rm{m}}\end{aligned}\)

The horizontal component (\(x\) component) of the resultant vector \({\rm{C}}\) is,

\({C_x} = {A_x} + {B_x}\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{C_x} &= \left( {74.69\;{\rm{m}}} \right) + \left( { - 20.03\;{\rm{m}}} \right)\\ &= 54.66\;{\rm{m}}\end{aligned}\)

The vertical component (\(y\) component) of the vector \({\rm{A}}\) is,

\({A_y} = - A\sin \left( {{{21}^ \circ }} \right)\)

Here \(A\) is the magnitude of the vector \({\rm{A}}\).

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{A_y} &= - \left( {80\;{\rm{m}}} \right) \times \sin \left( {{{21}^ \circ }} \right)\\ &= - 28.68\;{\rm{m}}\end{aligned}\)

The vertical component (\(y\) component) of the vector \({\rm{B}}\) is,

\({B_y} = B\cos \left( {{{11}^ \circ }} \right)\)

Here \(B\) is the magnitude of the vector \({\rm{B}}\).

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{B_y} &= \left( {105\;{\rm{m}}} \right) \times \cos \left( {{{11}^ \circ }} \right)\\ &= 103.07\;{\rm{m}}\end{aligned}\)

The vertical component (\(y\) component) of the resultant vector \({\rm{C}}\) is,

\({C_y} = {A_y} + {B_y}\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{C_y} &= \left( { - 28.68\;{\rm{m}}} \right) + \left( {103.07\;{\rm{m}}} \right)\\ &= 74.39\;{\rm{m}}\end{aligned}\)

The magnitude of the resultant vector \({\rm{C}}\) is,

\(C = \sqrt {C_x^2 + C_y^2} \)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}C &= \sqrt {{{\left( {54.66\;{\rm{m}}} \right)}^2} + {{\left( {74.39\;{\rm{m}}} \right)}^2}} \\ &= 92.3\;{\rm{m}}\end{aligned}\)

The direction of the resultant vector \({\rm{C}}\) is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{C_y}}}{{{C_x}}}} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}\theta &= {\tan ^{ - 1}}\left( {\frac{{74.39\;{\rm{m}}}}{{54.66\;{\rm{m}}}}} \right)\\ &= {53.7^ \circ }\end{aligned}\)

Hence, the length of the side \({\rm{C}}\)is \(92.3\;{\rm{m}}\), and its orientation is \({53.7^ \circ }\) south of west.