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So essentially, we have an aircraft $A$ moving true north ($0^oT$) at a velocity of $10kn$. The aircraft started at a position of $A(x,y)$. There are 2 radio stations transmitting the bearing to the aircraft (The bearing from the station to the aircraft). Station $P$ is at a location $P(10,10)$ and station $Q$ at $Q(10,2)$. The bearings are $90^o$ and $45^o$ respectively at $t=0$. Find an equation for the lines $PA$ and $QA$ so that at any time $t$ you can equate these equations to find the co-ordinates of the aircraft.

As seen below, the point $A(x,y)$ is the intersection of the 2 linear equations $PA$ and $QA$. Obviously the two equations will change relative to the movement of the aircraft's movement.

What I can't understand is how to make an equation for both lines that will work for all values of $t$. So that when $t=1$, I can have 2 equations with new angles etc that when equated can provide the location of $A$.

Also, the bearings when $t>0$ are unknown.

Any help is appreciated, this is only of personal interest.

Illustration of the described situation.

2 Answers 2

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The general equation of a line going threw $A(x,y)$ is $\alpha X+\beta Y=\alpha x+\beta y$. If such a line passes threw $P(10,10)$, we have $$10\alpha+10\beta=x\alpha+y\beta$$ Taking for example $\alpha=10-y$, you find $\beta=x-10$. So equation of line $(PA)$ is $$(10-y)(X-x)+(x-10)(Y-y)=0$$ You can do the same for line $(QA)$. Of course, the coefficients of your equations depend on the location of the plane :-)

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    Sorry, I'm confused. This would apply on one occasion while $t=0$ or are the coefficients related to the bearings? – 2017-02-16
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    They are correct for every position of the plane, but of course they vary according to this position :-) – 2017-02-16
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Alright I have solved my question. Using the equation $(y_A - y_P)=m(x_A-x_P)$ for both points $P(10,10)$ and $Q(10,2)$. Since the value of $x_A$ doesn't change (It's moving true north), I made an equation for $y_A$: $y_a=(x_A-x_P)tan(\theta_P)+y_P$ which equates to $y_A=8tan(\theta_Q)+10$. The tricky part however (not that tricky, really) is finding an expression for $\theta_Q$ since it changes relative to the movement of the aircraft.

I did this by making a triangle, the rise was $10t$, and the run $(18-x_P)$. So by using inverse sin we can find the angle (bearing) of $PA$.

$Sin^{-1}(\frac{10t}{\sqrt{8^2+(10t)^2}})$. Then since you can use $tan(\theta)$ to find the gradient of a line $m$, I substituted that into the place of $m$ in the original equation.

Therefore, the final equation is: $y_A=8tan(Sin^{-1}(\frac{10t}{\sqrt{8^2+(10t)^2}}))+10$. So for any time t, we have $A(y_A, 10t+10)$.