(a) 0.85 m
Let's start by analyzing the motion of beetle 1. We need to resolve its motion along two directions: east-west (x-direction) and north-south (y-direction).
[tex]A_x = 0.50 + 0.80 cos 30^{\circ} =1.19 m[/tex]
[tex]A_y = 0.80 sin 30^{\circ} =0.40 m[/tex]
Now let's do the same with beetle 2:
[tex]B_x = 1.60 sin 40^{\circ}+ x_2 = 1.03 + x_2[/tex]
[tex]B_y = 1.60 cos 40^{\circ} + y_2 = 1.23 + y_2[/tex]
where [tex]x_2,y_2[/tex] are the components of the second run of beetle 2, which are unknown.
We want beetle 2 to end up at the same location of beetle 1, so the x- and y- component of the displacement of the two beetles must be the same. Therefore:
[tex]A_x = B_x\\1.19 = 1.03 + x_2 \rightarrow x_2 = 1.19-1.03 = 0.16 m[/tex]
[tex]A_y = B_y \\0.40 = 1.23+ y_2 \rightarrow y_2 = 0.40-1.23 = -0.83 m[/tex]
So, the magnitude of the displacement of the second run of beetle 2 must be
[tex]d=\sqrt{x_2^2+y_2^2}=\sqrt{0.16^2+(-0.83)^2}=0.85 m[/tex]
(b) [tex]79.1^{\circ}[/tex] south of east
In order to find the direction of the second run of beetle 2, let's consider again the components of this displacement:
[tex]x_2 = 0.16 m\\y_2 =-0.83 m[/tex]
We can find the direction by using the formula
[tex]tan \theta = \frac{y_2}{x_2}[/tex]
By substituting,
[tex]tan \theta = \frac{-0.83}{0.16}=-5.188\\\theta = tan^{-1}(-5.188)=-79.1^{\circ}[/tex]
which means [tex]79.1^{\circ}[/tex] south of east.
