This question can be solved using the concept of the composition of the vector.
(a) The magnitude of F is b "164.27 N".
(b) The angle of the vector from horizontal is " ".
(a)
From the composition of the rectangular components of a vector, we know that the magnitude of the resultant vector is given by the following formula:
[tex]F = \sqrt{F_x^2+F_y^2}[/tex]
where,
[tex]F_x[/tex] = x-component of the force = 122 N
[tex]F_y[/tex] = y-component of the force = 110 N
Therefore,
[tex]F = \sqrt{(122\ N)^2+(110\ N)^2}[/tex]
F = 164.27 N
(b)
The angle of the vector is given by the following formula:
[tex]\theta = tan^{-1}(\frac{F_y}{F_x})\\\\\theta = tan^{-1}(\frac{110\ N}{122\ N})\\\\[/tex]
θ = 42.04°
Learn more about the composition of vector here:
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