Find the magnitude and direction of the resultant force in Red. One displacement is of $26km$ on a bearing of $175°$ and another one is $18km$ of a bearing of $294°$.
The bottom right angle of the triangle is $61°$, so I use cosine rule:
$R^2=26^2+18^2-(26)(18)cos(61)$
$R=27.8km$
Now to work out the direction using the sine rule:
$\dfrac{sin(x)}{18}=\dfrac{sin(61)}{27.8}$
$x=34.5°$
$175+34.5=209.5°$
Now, according to my book this is wrong, the actual values are:
$23.4km$
$217.3°$
Where am I going wrong?
You have a mistake using cosine law, it is : $$ C^2 = A^2 + B^2 - \color{Red}{2}AB\cos{\theta}. $$
If $\theta = 61^{\circ}$, $A = 26km$ and $B = 18km$, then $$ C = \sqrt{26^2 + 18^2 - \color{Red}{2}(26)(18)\cos{61^{\circ}}}km = 23.37km \approx \boxed{23.4km}. $$
For that, using sine law $$ \frac{\sin{\alpha}}{B} = \frac{\sin{\theta}}{C}, $$ you get $$ \alpha = \arcsin{\left(\frac{18}{23.4}\sin{61^{\circ}}\right)} = 42.28^{\circ} \approx 42.3^{\circ}. $$
The total angle is $175^{\circ} + 42.3^{\circ} = \boxed{217.3^{\circ}}.$