Determine the magnitude and the direction of the velocity of the aircraft. An aircraft A is flying in a vertical plane containing two tracking stations P and Q which are 15km apart. At a cetain instant θ (measured anticlockqise from horizontal line PQ to the line PA) is 60° and the $ \dot \theta $ (angular velocity) is -0.025rad/s. At the same time, α (measured anticlockqise from extended horizontal line PQ to the line QA) is 150° and the $\dot \alpha$ is -0.02rad/s. Determine the magnitude and the direction of the velocity of the aircraft.

I used the radial and transverse components of the velocities relative to P and Q and calculated the resultant. 

Relative to P, radial component = $$\frac{d(PA)}{dt}= \frac{d(15000Cos(\theta))}{dt} = -15000Sin(\theta) \dot \theta $$ 

The transverse component = $$r \dot \theta =-(PA)0.025$$

The components of velocity relative to Q could be obtained in the same way but I'm not sure about what to do next. Is taking the resultant of them correct? I took the resultant and got the following answers.

Magnitude - 15133m/s

Direction - 88.89° north of east
Consider the triangle $\Delta PQA$. Denote the measure of angle $\angle APQ$ as $\theta$ and note $\theta=\pi/3\;\text{rad}$, the measure of angle $\angle AQP$ as $\beta$ and note $\beta=\pi-\alpha$ where $\alpha$ is the measure of the angle between the extended line $QP$ and segment $AQ$. Since $\alpha=5\pi/6\;\text{rad}$ we know $\beta=\pi/6\;\text{rad}$. We know that $\theta' = -0.025\;\text{rad/s}$ and that $\alpha' = -0.02\;\text{rad/s}$. This implies that $\beta' = 0.02\;\text{rad/s}$.

Define our coordinates centered at point $P$ with the $x$ direction along segment $PQ$ and the $y$ direction perpendicular to segment $PQ$.

Let point $A=(x,y)$. The velocity of point $A$ (denoted $V_A$) is $(x',y')$. The magnitude of the velocity is $\sqrt{x'^2 + y'^2}$ and the direction (relative to the horizontal $PQ$) is $\tan{y'/x'}$. It's easy to show that $A=(x,y)=(3.75\;\text{km},6.495\;\text{km})$.

Denote the point $B$ as the vertical projection (perpendicular to $PQ$) of point $A$ onto segment $PQ$. Consider the triangle $\Delta PBA$. The triangle is a right triangle with a vertical side of length $y$ and a horizontal side of length $x$. Note that

$$\tan\theta = \frac{y}{x}$$

which can be derived to produce the line

$$y'=(\tan\theta) x' + x(\sec^2{\theta})\theta'$$

where all but $x'$ and $y'$ are known. Substituting all known values reduces the line to

$$y'=1.732x'-0.375\;\text{km/s}$$

Now consider triangle $\Delta QBA$. Again note that this is a right triangle with a vertical side of length $y$ and a horizontal side of length $15-x$. Again note that

$$\tan{\beta}=\frac{y}{15-x}$$

which can be derived to produce the line

$$y'=-(\tan{\beta})x'+(15-x)(\sec^2{\beta})\beta'$$

where again, all is known except for $x'$ and $y'$. Substituting in the known values reduces the line to

$$y' = -0.5774x'+0.3\;\text{km/s}$$

The intersection of these two lines will provide us with the velocity vector 

$$(x',y')=(0.292\;\text{km/s},0.131\;\text{km/s})$$

The magnitude, as described above is, $320\;\text{m/s}$ and the direction is $0.418\;\text{rad}=23.9^o$ above the horizontal. To be even more clear we will say that the ground projection of the plane is heading toward $Q$ and away from $P$.
$$OA = OPtan(\theta)$$
$$d(OA) = d(OP)tan(\theta) + OPsec^{2}(\theta)d\theta ......(1)$$

Similarly,

$$OA = OQtan(\alpha)$$
$$d(OA) = d(OQ)tan(\alpha) + OQsec^{2}(\alpha)d\alpha  ........(2)$$

OQ = 15000-OP

$$\frac{OA}{OP} = tan(60)$$

$$\frac{OA}{OQ} = tan(30) = \frac{OA}{15-OP} = tan(30)$$

Solving these two :  You will find that $OP = 3.75$$

$$ \frac{d(OA)}{dt} = \frac{d(OP)}{dt}tan(\theta) +OPsec^{2}(\theta)\frac{d\theta}{dt}...(3)$$ $$ \frac{d(OA)}{dt} = \frac{d(OQ)}{dt}tan(\alpha)+ OQsec^{2}(\alpha)\frac{d\alpha}{dt}.....(4)$$

$$\frac{d\theta}{dt} = \theta^{o}$$
$$\frac{d\alpha}{dt} = \alpha^{o}$$

Substituting the values of $\theta, \alpha,\theta^{o},\alpha^{o}, OP, OQ$,

we get,  and solving (3) and (4) we get the y component $\frac{d(OA)}{dt} $ and the x component $\frac{d(OP)}{dt}$ and take the magnitude $\sqrt{x^2+y^2}$ou will get 

$$ v = 320$$m/sec and the angle will be the $tan^{-}(\frac{y}{x}) = 23.9$ degrees.