CAIE A-Level Mathematics AS 4.1.2 The Vector Nature of Force Questions

$F=20 \sin 60$
Attempt to resolve in y-direction;
2 terms;
must be $20 \cos 60$ or $20 \sin 60$
and must be linked to F (can be implied by the correct answer seen only).

$=17.3 \mathrm{~N}$
AWRT 17.3 (17.320508…) or $10 \sqrt{3}$.

For resolving in any direction

*M1

Correct number of terms;
allow sin/cos mix;
allow sign errors.

$$\begin{aligned} (\text { Horizontal component }=X=R \cos \theta)= \pm(20 \cos 60)[ \pm 10] (\text { Vertical component }=Y=R \sin \theta)= \pm(20 \sin 60-10) {[= \pm 7.3205 \ldots]} \end{aligned}$$
For both correct.

$$\begin{aligned} \text { Magnitude }=\sqrt{(20 \sin 60-10)^{2}+(20 \cos 60)^{2}} {[=12.393136 \ldots]} \end{aligned}$$
OE - correct number of terms.

Angle $=\tan ^{-1}\left(\frac{20 \sin 60-10}{20 \cos 60}\right)[=36.206023 \ldots]$
OE (e.g. reciprocal) - correct number of terms.

Magnitude $=12.4 \mathrm{~N}$
and Direction $=36.2^{\circ}$ above (positive) x-axis
OE for direction e.g. $36.2^{\circ}$ anticlockwise from (positive) x-direction, $36.2^{\circ}$ above the horizontal.

Possibly seen on a diagram.

(Radians: 0.63191... to 3sf or better)