2 \includegraphics[max width=\textwidth, alt={}, center]{f4f2996b-5382-4b0d-9804-b5f5945946b3-2_636_519_664_813} Three horizontal forces of magnitudes \(F \mathrm {~N} , 63 \mathrm {~N}\) and 25 N act at \(O\), the origin of the \(x\)-axis and \(y\)-axis. The forces are in equilibrium. The force of magnitude \(F \mathrm {~N}\) makes an angle \(\theta\) anticlockwise with the positive \(x\)-axis. The force of magnitude 63 N acts along the negative \(y\)-axis. The force of magnitude 25 N acts at \(\tan ^ { - 1 } 0.75\) clockwise from the negative \(x\)-axis (see diagram). Find the value of \(F\) and the value of \(\tan \theta\).

## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving components of $F$ in $x$ and $y$ directions |
| $F_x = F\cos\theta = 25 \times 0.8 = 20$, $F_y = F\sin\theta = 63 - 25 \times 0.6 = 48$ | A1 | |
| | M1 | For using $F = \sqrt{F_x^2 + F_y^2}$ or for using $\tan\theta = F_y \div F_x$ |
| $F = 52$ N or $\tan\theta = 2.4$ | A1 | |
| $\tan\theta = 2.4$ or $F = 52$ N | B1 | |
| **Total: 5 marks** | | |

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\includegraphics[max width=\textwidth, alt={}, center]{f4f2996b-5382-4b0d-9804-b5f5945946b3-2_636_519_664_813}

Three horizontal forces of magnitudes $F \mathrm {~N} , 63 \mathrm {~N}$ and 25 N act at $O$, the origin of the $x$-axis and $y$-axis. The forces are in equilibrium. The force of magnitude $F \mathrm {~N}$ makes an angle $\theta$ anticlockwise with the positive $x$-axis. The force of magnitude 63 N acts along the negative $y$-axis. The force of magnitude 25 N acts at $\tan ^ { - 1 } 0.75$ clockwise from the negative $x$-axis (see diagram). Find the value of $F$ and the value of $\tan \theta$.

\hfill \mbox{\textit{CAIE M1 2015 Q2 [5]}}