Law of Cosines Geometry

Question

Included below is a picture of a geometry question I’ve been tinkering with. I feel like this should involve law of cosines to begin placing the angles, but I’m not sure how to proceed.

Answer 1

No information about the location of $E$ is given except that its distance from $D$ is $6.$ So let $E$ move all the way around the circle of radius $6$ centered at $D.$ Ask whether the distance $EF$ varies as $E$ moves along that circle. Clearly it does, but all of the specified information about the figure remains as it was; it doesn't change. Therefore the given information is not enough to say what the distance $FE$ is. But it cannot be more than $12+8+6.$

Answer 2

You can get all the needed info like this, if angles ABC and CDE are right:

$\tan(ABC) = AB/BC\\ AC = \sqrt{AB^2+BC^2}$

Since you now know AC, CF, and FA, you can get $\cos(ACF)$ from the law of cosines.

$\tan(ECD) = ED/DC\\ CE = \sqrt{CD^2+DE^2}$

Since $ACB+ACF+FCE+ECD = \pi$, you can now get $FCE$.

Since you now know $FCE, FC, CE$, you can get $EF$ from the law of cosines.

Note that $\tan(x) =\dfrac{\sin(x)}{\cos(x)} =\dfrac{\sqrt{1-\cos^2(x)}}{\cos(x)} $ so $\cos^2(x)\tan^2(x) =1-\cos^2(x) $ so $\cos^2(x)(\tan^2(x)+1) =1 $ or $\cos(x) =\dfrac1{\sqrt{\tan^2(x)+1}} $.

Also note that $\arctan(x)+\arctan(y) =\arctan\left(\dfrac{x+y}{1-xy}\right) $.

Answer 3

See Figure

Since angle ABC and CDE are not given therefore there are infinite no. of solutions for this problem. Minute change in angle will cause change in length of FE itself.