Pearson Edexcel IAL Mathematics M1.5.1 Forces treated as vectors

R=F+G

$$\begin{array}{ll} R^{2}= 8^{2}+10^{2}-2 \times 8 \times 10 \cos 120^{\circ} \text { oe }(244) \text { OR: } R^{2}=8^{2}+10^{2}+2 \times 8 \times 10 \cos 60^{\circ} \text { OR: } R^{2}=(8 \sin 60)^{2}+\left(10+8 \cos 60^{\circ}\right)^{2} \text { OR: } R^{2}=(10 \sin 60)^{2}+\left(8+10 \cos 60^{\circ}\right)^{2} \end{array}$$

M1A1

$R=\sqrt{244}=15.620499 \ldots \mathrm{~N}$

A1

$$\begin{aligned} \frac{\sin \alpha}{8}=\frac{\sin 120^{\circ}}{\sqrt{244}}\left(\text { allow } \sin 60^{\circ}\right) \text { OR } \frac{\sin \beta}{10}=\frac{\sin 120^{\circ}}{\sqrt{244}}\left(\text { allow } \sin 60^{\circ}\right) \text { OR } 8^{2}=(\sqrt{244})^{2}+10^{2}-2 \times \sqrt{244} \times 10 \cos \alpha \text { OR } 10^{2}=(\sqrt{244})^{2}+8^{2}-2 \times \sqrt{244} \times 8 \cos \beta \text { OR } \tan \alpha=\frac{8 \sin 60}{10+8 \cos 60^{\circ}} \text { or } \sin \alpha=\frac{8 \sin 60}{\sqrt{244}} \text { or } \cos \alpha=\frac{10+8 \cos 60^{\circ}}{\sqrt{244}} \text { (or reciprocal of tan) } \text { OR } \tan \beta=\frac{10 \sin 60}{8+10 \cos 60^{\circ}} \text { or } \sin \beta=\frac{10 \sin 60}{\sqrt{244}} \text { or } \cos \beta=\frac{8+10 \cos 60^{\circ}}{\sqrt{244}} \text { (or reciprocal of tan) } \end{aligned}$$

M1A1

$\alpha=26 .(3 . .)^{\mathrm{o}} \quad$ OR $\beta=33.67 \ldots{ }^{\mathrm{o}}$ (accept 34)

A1

Bearing is $206^{\circ}$ (nearest degree)

A1

(7)

Notes for question 5