Note that $\theta_{out}$ has to depend also on sizes of legs and turns (may be on their ratio). That is because on the outbound leg, you compensate for drift the wind causes on the turns.
In case of instant turns, the outbound leg is identical to the inbound leg (opposite direction), so $\theta_{out}=\theta_{in}$
The case of very long legs is similar to this: $\theta_{out}=\theta_{in}$ approximately.
Edit:
We will use
- length unit equal to 1 minute of flight
- Ground coordinate system (GCS)
- Air coordinate system (ACS)
- wind speed - vector $(w_1, w_2)$
The plane moves in ACS in any direction by a fixed speed v=1 unit per minute (so lengths can be given in minutes).
The trajectory in GCS is pictured in the question.
The trajectory in ACS consists of
- segment of angle $\theta_{in}$ southwards, length 1 minute,
- then circle arc of length $t_{Tout}$, [ T stands for Turn ]
- second segment of angle $\theta_{out}$ (unknown), also southwards, length $t_{Lout}$ (unknown), [ L stands for Leg ]
- and second circle arc length $t_{Tin}$.
$\theta_{in}$ actually corresponds to the wind strength and direction. I think
$w_2=\sin \theta_{in}$ which can be substituted in final formula.
Times $t_{Tin}$ and $t_{Tout}$ are approximately 1 minute. Precisely they are determined
by the angles (I think $t_{Tout} = 1 - (\theta_{in} + \theta_{out}) / 180$ and
$t_{Tin} = 1 + (\theta_{in} + \theta_{out})/180$).
Important is that the two arcs make together the whole circle.
That also implies $t_{in}+t_{out}= 2 $ minutes.
The movements in ACS in changed order are:
- two arcs, together they make no move (but it takes 2 minutes of time)
- first segment (corresponds to inbound leg) - 1 minute, vector $(\cos \theta_{in}, \sin \theta_{in})$
- second segment (corresponds to outbound leg).
In GCS, we have to add to this:
- drift caused by wind during the circle and first segment. This is 3 minutes, hence $(3w_1, 3w_2)$
- drift during the second segment.
The circle plays no role for calculations. This is good news. Things about segments are easier to calculate.
The movement in GCS together, with exception of second segment, can be read from the the above, and it is $(3w_1 + \cos \theta_{in}, 3w_2 - \sin \theta_{in})$.
This is the vector we have to travel (in the opposite direction) during the second segment. Indeed we have to be on the same point in GCS in the start and end of the maneavour. That means, we have already the direction of the second segment in GCS:
$$
\alpha_{out} = \arctan \frac{ 3w_2 - \sin \theta_{in} }{ 3w_1 + \cos \theta_{in} }
$$
Note this is not $\theta_{out}$, but it determines it.
Certainly the pilot is able to fly in the direction $\alpha_{out}$,
and he is able to determine corresponding $\theta_{out}$. I am lazy
to do the calculation of the corresponding $\theta_{out}$, and I'm not sure its possible by elementary functions.
As regards the time, the calculation is now straightforward.
The vector speed in GCS is $( - \cos \theta_{out} + w_1, - \sin \theta_{out} + w_2 )$
so the time needed is
$$
t_{Lout} = \frac { 3w_1 + \cos \theta_{in} }{ - \cos \theta_{out} + w_1 }
= - \frac { 3w_2 - \sin \theta_{in} }{ - \sin \theta_{out} + w_2 }
$$
Please check my calculations. Especially the signs.
Edit: Nota bene, for small winds we get approximately $\theta_{in}=w_2$ and
$
\theta_{out} = \alpha_{out} = 2 \theta_{in}
$.
The blue arrow show the two stages of where the camel walked. The red arrow passes from where it wound up to Oasis B.
North is at the top of this graph. When the camel was facing South (toward the bottom of the graph), "West of South" would be 15º to the camel's right, since West is to the right when facing South (that is leftward from bottom on the graph $^*$). It then turned around and traveled North (back toward the top of the graph).
$^*$ The graph is an overhead view of these proceedings...
What the camel still needs to do is journey along the red vector. Part (a) asks how long this vector is and Part (b) asks what angle the vector makes above (positive) or below (negative) going straight to the right on the graph, that is, due East (the 0º reference direction).
Best Answer
To the problem with vectors, we have to break the trip up into two parts, I'll call them $A$ (airport to change heading) and $B$ (change heading to emergency landing) like you did. From the picture we can see that the vector $A$ uses an angle of $68^{\circ}$ from the vertical. So $A = \langle 140\sin(68^{\circ}), 140\cos(68^{\circ})\rangle \approx \langle 129.806, 52.445\rangle$. You could also have done $A = \langle 140\cos(22^{\circ}), 140\sin(22^{\circ})\rangle$ if you wanted to work with the angle formed between $A$ and the horizontal. For vector $B$, we have the angle with the horizontal but it is below the horizontal. So the vertical component of $B$ is negative. $B = \langle 255\cos(48^{\circ}), -255\sin(48^{\circ})\rangle \approx \langle 170.628, -189.502\rangle$. The resultant is $R = A+B = \langle 300.434, -137.507\rangle$. $|R| = \sqrt{(300.434)^2+(-137.507)^2} \approx 330.220$ km. The angle they need to travel at is $\tan^{-1}\left(\frac{137.507}{300.434}\right) \approx 27.24^{\circ}$ below the horizontal or south of east.