Step 1

We are given the force acting along the cable B is ${F_B} = 55\;{\rm{N}}$.

We are given the force acting along the cable C is ${F_C} = 40\;{\rm{N}}$.

We are asked the angles $\theta $ and $\phi $ made between the axes OA of the flag pole and AB and AC, respectively, of each cable.

 
Step 2

The following is the diagram of the system:

We have the vector representation of point O is $O = \left( {0i + 0j + 0k} \right){\rm{ m}}$.

We have the vector representation of point A is $A = \left( {0i + 4j + 3k} \right){\rm{ m}}$.

We have the vector representation of point B is $B = \left( {1.5i + 0j + 6k} \right){\rm{ m}}$.

We have the vector representation of point C is $C = \left( { - 2i + 0j + 4k} \right){\rm{ m}}$.

 
Step 3

The magnitude of position vector of AO is,

\[{R_{AO}} = O - A\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{l} {R_{AO}} = \left( {0i + 0j + 0k} \right){\rm{ m}} - \left( {0i + 4j + 3k} \right){\rm{ m}}\\ {R_{AO}} = \left( {0i - 4j - 3k} \right){\rm{ m}} \end{array}\]
 
Step 5

The magnitude of position vector of AB is,

\[{R_{AB}} = B - A\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{l} {R_{AB}} = \left( {1.5i + 0j + 6k} \right){\rm{ m}} - \left( {0i + 4j + 3k} \right){\rm{ m}}\\ {R_{AB}} = \left( {1.5i - 4j + 3k} \right){\rm{ m}} \end{array}\]
 
Step 7

The magnitude of position vector of AC is,

\[{R_{AC}} = C - A\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{l} {R_{AC}} = \left( { - 2i + 0j + 4k} \right){\rm{ m}} - \left( {0i + 4j + 3k} \right){\rm{ m}}\\ {R_{AC}} = \left( { - 2i - 4j + 1k} \right){\rm{ m}} \end{array}\]
 
Step 9

The magnitude of angle $\theta $ made between the axes OA of the flag pole and AB is,

\[\cos \theta = \frac{{{R_{AO}} \cdot {R_{AB}}}}{{\left| {{R_{AO}}} \right| \cdot \left| {{R_{AB}}} \right|}}\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{c} \cos \theta = \frac{{\left( {0i - 4j - 3k} \right){\rm{ m}} \cdot \left( {1.5i - 4j + 3k} \right){\rm{ m}}}}{{\sqrt {{{\left( 0 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 3} \right)}^2}} \;{\rm{m}} \cdot \sqrt {{{\left( {1.5} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 3 \right)}^2}} \;{\rm{m}}}}\\ \cos \theta = \frac{{\left( 0 \right) + \left( {16} \right) + \left( { - 9} \right)}}{{\sqrt {25} \sqrt {27.25} }}\\ \cos \theta = 0.27\\ \theta = 74.4^\circ \end{array}\]
 
Step 11

The magnitude of angle $\phi $ made between the axes OA of the flag pole and AC is,

\[\cos \phi = \frac{{{R_{AO}} \cdot {R_{AC}}}}{{\left| {{R_{AO}}} \right| \cdot \left| {{R_{AC}}} \right|}}\]
 
Step 12

Substitute the values in the above expression.

\[\begin{array}{c} \cos \phi = \frac{{\left( {0i - 4j - 3k} \right){\rm{ m}} \cdot \left( { - 2i - 4j + 1k} \right){\rm{ m}}}}{{\sqrt {{{\left( 0 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 3} \right)}^2}} \;{\rm{m}} \cdot \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 1 \right)}^2}} \;{\rm{m}}}}\\ \cos \phi = \frac{{\left( 0 \right) + \left( {16} \right) + \left( { - 3} \right)}}{{\sqrt {25} \sqrt {21} }}\\ \cos \phi = 0.567\\ \phi = 55.4^\circ \end{array}\]