Answer
$0.84\ m$
Work Step by Step
Let the origin be the mutual starting point,
and let east be the +x direction of our coordinate system.
Angles in the formulas are relative to the +x axis (counterclockwise=positive).
$30^{o}$ north of east $\Rightarrow 30^{o}$ relative to +x,
$40^{o}$ east of north $\Rightarrow 50^{o}$ relative to +x.
BEETLE 1.
First leg: $\vec{a}=(0.50m)\hat{i}$
Second leg: $\vec{b}=(0.80m)(\cos 30^{o})\hat{i}+(0.80m)(\sin 30^{o})\hat{j}\qquad $... using (3-5)
Final position: $\vec{a}+ \vec{b}=[0.50m+(0.80m)(\cos 30^{o})]\hat{i}+(0.80m)(\sin 30^{o})\hat{j}$
BEETLE $2$.
First leg: $\vec{c}=(1.60m)(\cos 50^{o})\hat{i}+(1.60m)(\sin 50^{o})\hat{j}\qquad $... using (3-5)
Second leg: $\vec{d}=x\hat{i}+y\hat{j}$
Final position: $\vec{c}+ \vec{d}=[(1.60m)(\cos 50^{o})+x]\hat{i}+[(1.60m)(\sin 50^{o})+y]\hat{j}$
Both beetles should have the same the same final position, so the components must be equal.
$\left[\begin{array}{llll}
0.50m+(0.80m)(\cos 30^{o}) & = & (1.60m)(\cos 50^{o})+x & (1)\\
(0.80m)(\sin 30^{o}) & = & (1.60m)(\sin 50^{o})+y & (2)
\end{array}\right]$
Solving equation (1)
$x=0.50m+(0.80m)(\cos 30^{o})-(1.60m)(\cos 50^{o})=0.16 m$
Solving equation ($2$)
$y=(0.80m)(\sin 30^{o})-(1.60m)(\sin 50^{o})=-0.83m$
The second leg of beetle 2:
$\vec{d}=(0.16 m)\hat{i}+(-0.83m)\hat{j}$
$(a)$
The magnitude of $\vec{d}$ is
$d=\sqrt{(0.16 m)^{2}+(-0.83m)^{2}}=0.84m$
