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

geometrytriangles

In this triangle

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?

Best Answer

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.

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