Three forces F₁, F₂ and F₃ acting on a particle P are given by F₁ = (7i – 9j) N F₂ = (5i + 6j) N F₃ = (pi + qj) N where p and q are constants - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1

After removing force F₁, we need to find resultant R from forces F₂ and F₃.

F₂ is given as (5i + 6j) N and F₃ as (pi + qj) N, substituting the values of p and q:

F₃ = (-12i + 3j) N.

Now, combine F₂ and F₃:

R = (5i + 6j) + (-12i + 3j)
= (-7i + 9j) N.

To find the magnitude of R:

$|R| = \\sqrt{(-7)^2 + (9)^2} = \\sqrt{49 + 81} = \\sqrt{130} \\approx 11.4 ext{ N}$

To find the angle θ that the resultant R makes with the j-axis, we can use the tangent function:

$\tan(θ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{|-7|}{9}$

Solving for θ:

$θ = \tan^{-1}\left(\frac{7}{9}\right) \\approx 39.8^{\circ}$

However, since we want the angle to the nearest degree, we round it:

$θ \approx 40^{\circ}$