Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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?
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.
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.
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.
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.
