I am given 2 side lengths for one triangle and two side lengths for the parallelogram. I am asked to find the length of m (FE) and n (DE)
I am given the lenghts:
I don't see how to use Law of Sines because I don't have any angles and I don't see how to use Law of Cosines to solve triangle ACF because I am missing a side length.
Am I missing a concept or is this problem missing given information?
On
welcome to stack exchange. The answer will simply be that $m = 15$ and $n = 9$. Lets say you made a new point G opposite E then joined C to G and then G to E, you'd effectively end up with the same parallelogram as you have underneath.
On
Note that the triangles $AFC$ and $EFD$ are similar. Hence $\frac{n}{m}=\frac{9}{15}$.
Now let the angle at E in $EFD$ be $x$. Then the angle at $A$ in $AFC$ is also $x$ by similarity. Using the cosine rule you can now find the lenght of g as a function of $x$, i.e. $g(x)$. Then $l(x)=16-g(x)$.
We have $n=\frac{9}{15}m$, hence we can now find $n(x)$ using the cosine rule in $EFD$. This thus gives $m(x)=\frac{15}{9}n(x)$. Now look at triangle $ABE$. Note that we know all sides as a function of $x$ and know that its angle at $E$ is $x$. Hence we can solve for $x$ by applying the cosine rule.
I however suspect that the actual calculations will be very ugly, so there probably is another solution possible.
On
Well, you can certainly reconstruct the image by leaning the parallelogram to different angles and sliding F to where it needs to be. In other words by adjusting $\angle CAB$ to different values.
The question is is there some reason that for all measures of $\angle CAB$ will the lines $FE$ And $ED$ should be the same for all such parallelograms. Off hand there is no reason they should.
But the more we increase $\angle CAB$ the more point $F$ will move to the right. And as $9 < 15$ there is some angle where $F$ will move so far to the right so as to superimpose itself on $D$. As a result $E$ will superimpose on $D$. Then $FE= ED =0$.
Therefore $FE$ and $DE$ are not constant and we don't have enough information.
I don't believe there's enough information. Let's let
$$A=(0,0)$$
$$B=(16,0)$$
$$C=(9\cos\theta,9\sin\theta)\text{ for some }0<\theta<90^{\circ}$$
$$D=B+C=(16+9\cos\theta,9\sin\theta)$$
$$F=\left(\sqrt{15^2-9^2\sin^2\theta},9\sin\theta\right)$$
Then we have a parallelogram $ABDC$ with a point $F$ on $CD$, such that $|AC|=9$, $|AF|=15$, and $|AB|=16$. Our next step is to find $E$. We'll be done if we can show that $|AE|$ is a non-constant function of $\theta$.
$E$ is the intersection of the lines determined by segments $AF$ and $BD$. To find the coordinates of $E$, we'll first find the equations of these lines. Using the point-slope form, we have that the equation for the line determined by segment $AF$ is:
$$y=\frac{9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}}\cdot x$$
Using the point-slope form, we have that the equation for the line determined by segment $BD$ is:
$$y=\frac{9\sin\theta}{9\cos\theta}\cdot (x-16)$$
Hence we can find the $x$-coordinate of $E$ by solving
$$\frac{9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}}\cdot x=\frac{9\sin\theta}{9\cos\theta}\cdot (x-16)$$
This gives us that
$$x=\frac{16\cdot\sqrt{15^2-9^2\sin^2\theta}}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
We can then plug this into the equation for the line determined by segment $AF$ to obtain that
$$y=\frac{16\cdot9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
Hence
$$E=\left(\frac{16\cdot\sqrt{15^2-9^2\sin^2\theta}}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta},\frac{16\cdot9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}\right)$$
It follows that
$$|AE|=\frac{16\cdot15}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
Note that if $\theta=30^{\circ}$, then $|AE|\approx36.8$, but if $\theta=60^{\circ}$, then $|AE|\approx28.9$. So $|AE|$ is a non-constant function of $\theta$.
Finally, note that $m=|AE|-15$. So $m$ is a non-constant function of $\theta$. We need more information.