# Can a non-right triangle be solved with only two sides if it is inside a parallelogram?

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:

• h (AC) = 9
• k (AF) = 15
• f (AB) = 16

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?

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• I think there needs to be more information to uniquely determine $m$ and $n$. I posted an explanation below. Please let me know if you have any questions. – Andrew Ostergaard Jan 15 at 1:17
• Are you sure you need to find the values of each of $m$ and $n$ individually? You can find their ratio by similar triangles, though where $f$ comes into it is a mystery. – David K Jan 15 at 1:30
• I fixed a couple of typos in my solution. Was anything unclear? Please let me know. I'll be happy to help. – Andrew Ostergaard Jan 15 at 1:41
• Oh, it's pretty clear. Thanks. --...David, yes. The problem I was shown is asking for those, but either way, can we really find the ratios by similar triangles on this problem? I think the problem is still missing data. I think that to find the ratios that way, the FED triangle would need to have at least one side defined. – Altarith Jan 15 at 1:53
• This question does have an answer I think. The idea is that after you introduce a variable, everything is expressible in terms of that variable. You can then apply the cosine/sine rule to any of the triangles in the figure to solve for that variable and obtain the result. Or if you have not used the similarity of the triangles, you can use that to solve for the variable. – Grimp0w Jan 15 at 2:46

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.

• I see. I also thought there was not enough information, but I wasn't sure about it. Thanks for the details. – Altarith Jan 15 at 1:17
• Can't you express both $m$ and $n$ in terms of $\theta$ and solve for $\theta$ by using that you know their quotient? – Grimp0w Jan 15 at 2:48
• The formula for $m$ is a bit complicated, but we've shown that $m=15\cdot f(\theta)$, where $f$ is a non-constant function of $\theta$. If we try to find a formula for $n$, we're going to end up with $n=9\cdot f(\theta)$. So $\frac{m}{n}=\frac{15}{9}$, but this won't help us find $\theta$. – Andrew Ostergaard Jan 15 at 4:20
• Here's how we know we can't solve for $\theta$: if it was possible to somehow solve for $\theta$, then what that would show is that there is only one value of $\theta$ for which the above choices of the points $A$, $B$, $C$, $D$, $E$, $F$ satisfy the given conditions. But for $0<\theta<90^{\circ}$, the above choices of the points will satisfy the given conditions regardless of what $\theta$ is, as you can check. – Andrew Ostergaard Jan 15 at 4:21

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.

• Were we given that angle $CAB$ is acute? – Andrew Ostergaard Jan 15 at 14:02
• "Were we given that angle CAB is acute? " No. Does that matter. – fleablood Jan 15 at 17:12
• We were'nt given that $CAB$ is acute but if $CF = g < 16$ if it is acute $CF= g=16$ if is right. and $CF=g > 16$ if it is obtuse. The image can not be drawn unless it is acute. – fleablood Jan 15 at 17:32

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.

• Did you prove that $m=15$? If the diagram you draw has a lot of symmetry, then it will look as though $m=15$, but we would need to prove this. – Andrew Ostergaard Jan 15 at 1:13
• Hi Yung. Thanks for the answer. I tried using Geogebra to intersect lines with the parallelogram and while the diagram I have shows that it's almost the same, when I intersect lines and measure, the measurements are different. If I use different measurements, the k line length changes, but so does the m line length which ends up being different. Is there a math logic behind that can show they would always be the same regardless of the angle the k line takes? – Altarith Jan 15 at 1:25

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.