91Ó°ÊÓ

The displacement along west of north is \(A = 12\;{\rm{m}}\)

The direction of displacement along west of north is\(\alpha = 20^\circ \).

The displacement along south of west is\(B = 20\;{\rm{m}}\)

The direction of displacement along south of west is\(\beta = 40^\circ \).

The displacement is resolved along horizontal and vertical direction then sum of all the horizontal components gives horizontal component of net displacement. The sum of vertical components gives vertical component of net displacement.

The horizontal component of the net displacement is given as:

\(\begin{align}{}{R_x} &= - A\cos \left( {90^\circ - \alpha } \right) - B\cos \beta \\{R_x} &= - \left( {A\sin \alpha + B\cos \beta } \right)\end{align}\)

Substitute all the values in the above equation.

\(\begin{align}{}{R_x} &= - \left( {\left( {12\;{\rm{m}}} \right)\left( {\sin 20^\circ } \right) + \left( {20\;{\rm{m}}} \right)\left( {\cos 40^\circ } \right)} \right)\\{R_x} &= - 19.43\;{\rm{m}}\end{align}\)

The vertical component of the net displacement is given as:

\(\begin{align}{}{R_y} &= A\sin \left( {90^\circ - \alpha } \right) - B\sin \beta \\{R_y} &= A\cos \alpha - B\sin \beta \end{align}\)

Substitute all the values in the above equation.

\(\begin{align}{}{R_y} &= \left( {12\;{\rm{m}}} \right)\left( {\cos 20^\circ } \right) - \left( {20\;{\rm{m}}} \right)\left( {\sin 40^\circ } \right)\\{R_y} &= - 1.58\;{\rm{m}}\end{align}\)

The distance from starting point is given as:

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

Substitute all the values in the above equation.

\(\begin{align}{}R &= \sqrt {{{\left( { - 19.43\;{\rm{m}}} \right)}^2} + {{\left( { - 1.58\;{\rm{m}}} \right)}^2}} \\R &= 19.5\;{\rm{m}}\end{align}\)

The direction of compass from starting point is given as:

\(\tan \theta = \frac{{{R_y}}}{{{R_x}}}\)

Substitute all the values in the above equation.

\(\begin{align}{}\tan \theta = \frac{{ - 1.58\;{\rm{m}}}}{{ - 19.43\;{\rm{m}}}}\\\theta = 4.65^\circ \end{align}\)

Therefore, you are at \(19.5\;{\rm{m}}\) from the starting point and direction of compass is \(4.65^\circ \) from west of south.