Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.

Vectors are physical quantities that have magnitude and direction.

The two vectors cannot be added by using simple algebraic rules. They can be added by using the triangle law of vector addition.

The vectors \({\rm{A}}\) and \({\rm{B}}\) are represented as,

Representation of vectors \({\rm{A}}\) and \({\rm{B}}\)

• The magnitude of the vector \({\rm{A}}\)is, \(A = 27.5\;{\rm{m}}\).
• The magnitude of the vector \({\rm{B}}\)is, \(B = 30\;{\rm{m}}\).

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

\({A_x} = {\rm{A}}\cos \left( {{{66}^ \circ }} \right)\)

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

\(\begin{aligned}{}{A_x} &= \left( {27.5\;{\rm{m}}} \right) \times \cos \left( {{{66}^ \circ }} \right)\\ &= 11.185\;{\rm{m}}\end{aligned}\)

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

\({B_x} = {\rm{B}}\cos \left( {{{112}^ \circ }} \right)\)

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

\(\begin{aligned}{}{B_x} &= \left( {30\;{\rm{m}}} \right) \times \cos \left( {{{112}^ \circ }} \right)\\ &= - 11.238\;{\rm{m}}\end{aligned}\)

The resultant of horizontal components vectors \({\rm{A}}\) and \({\rm{B}}\) is,

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

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

\(\begin{aligned}{}{R_x} &= \left( {11.185\;{\rm{m}}} \right) + \left( { - 11.238\;{\rm{m}}} \right)\\ &= - 0.053\;{\rm{m}}\end{aligned}\)

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

\({A_y} = A\sin \left( {{{66}^ \circ }} \right)\)

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

\(\begin{aligned}{}{A_y} &= \left( {27.5\;{\rm{m}}} \right) \times \sin \left( {{{66}^ \circ }} \right)\\ &= 25.123\;{\rm{m}}\end{aligned}\)

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

\({B_y} = B\sin \left( {{{112}^ \circ }} \right)\)

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

\(\begin{aligned}{}{B_y} &= \left( {30\;{\rm{m}}} \right) \times \sin \left( {{{112}^ \circ }} \right)\\ &= 27.816\;{\rm{m}}\end{aligned}\)

The resultant of vertical components vectors \({\rm{A}}\) and \({\rm{B}}\) is,

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

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

\(\begin{aligned}{}{R_y} &= \left( {25.123\;{\rm{m}}} \right) + \left( {27.816\;{\rm{m}}} \right)\\ &= 52.939\;{\rm{m}}\end{aligned}\)

The magnitude of the resultant vector is,

\(R = \sqrt {R_x^2 + R_y^2} \)

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

\(\begin{aligned}{}R &= \sqrt {{{\left( { - 0.053\;{\rm{m}}} \right)}^2} + {{\left( {52.939\;{\rm{m}}} \right)}^2}} \\ \approx 52.94\;{\rm{m}}\end{aligned}\)

The direction of the resultant vector is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{R_y}}}{{{R_x}}}} \right)\)

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

\(\begin{aligned}{}\theta &= {\tan ^{ - 1}}\left( {\frac{{52.939\;{\rm{m}}}}{{ - 0.053\;{\rm{m}}}}} \right)\\ &= - {89.9^ \circ }\end{aligned}\)

Hence, the location of the dock is \(52.94\;{\rm{m}}\), \({89.9^ \circ }\) north of west.