Consider the three right triangles in this diagram. The values that cannot change are:

AB=DE; AD=8; CG=17; DF=24.5; angle EDF>0; BC>0

Segment BC will vary. For any given length of BC, I need to be able to calculate the length of FG. How do I do that? I have a feeling the answer will be trig-based, but I can't come up with it. Thanks in advance.

enter image description here

For some background, the application of this specific problem is in Forestry. There is a simple tool called a Biltmore stick used for measuring tree diameter. The user holds the stick flush against the tree, aligning the left edge of the stick with the visual of the left edge of the tree, then without moving his/her eye, uses the marked graduations to read the diameter that aligns with the right visual edge of the tree. The distance between graduation marks decreases with increasing diameter, allowing for diameter to be read directly with a single measurement.

The formula for calculating the length to each graduation mark is flawed (even though it's been unchanged since the 1890's). The formula is based on a fixed distance between eye and stick, then uses the radius of the tree to determine the center point of the tree. However, because the measurement is taken at 4.5' (below eye level for most adults), there is a downward angle to the line created between the eye and stick, and therefore the distance to the center of the tree along that line is longer than the radius of the trunk (the hypotenuese of the triangle rather than the adjacent side).

I've drawn over the first diagram to show how the triangles given illustrate the application of the problem enter image description here

Point A is the eye; D is the pivot point of the shoulder; CG is the distance between eye level and 4.5'; DF is the length of my arm when holding a stick; CG is the center of the tree; BF is the close edge of the tree.

  • $\begingroup$ I cannot post an answer. With $ ( L=24.5, H=17,k=8 , BC=p, FG=q) \, $ a sixth order implicit dependence is obtained by elimination: $$ 4 H^6 p^2+8 H^5 k p^2+16 H^3 k p^2 (p^2-q2)+8 H k (p^3-p q2)^2+H^4 (4 p^2 (k^2-L2+3 p^2)-12 p^2 q2+q2^2)+2 H^2 (p^2-q2) (2 p^2 (k^2-2 L2+3 p^2)+(k^2+L2-6 p^2) q2+q2^2)+(p^2-q2)^2 (k^4+2 k^2 (-L2+q2)+(L2-2 p^2+q2)^2)=0 $$ $\endgroup$
    – Narasimham
    Aug 13 '19 at 5:30

One way to calculate to $FG$ is to calculate $AG - AF$, the value of $AG$ can be expressed as a function of $BC$ (use the Pythagorean theorem), and in order to calculate $AF$, you will need the length of $EF$, again, use the Pythagorean theorem. Can you take it from there?

  • $\begingroup$ The length of EF will vary dependent on the length of BC. Is there a way to express EF as a function of BC? $\endgroup$
    – matt
    Aug 12 '19 at 20:33
  • $\begingroup$ $EF^2 + DE^2 = DF^2$, then $EF^2 = 24.5^2 - DE^2$ $\endgroup$ Aug 12 '19 at 21:20

So $\angle EDF$ is fixed but we don't know what it is. So we have the angles and the hypotenuse of $\triangle DEF$ so every component of $\triangle DEF$ is fixed.

$AD= BE=8$ is fixed so $BF = BE+EF$ is fixed and $AB=AD$ is fixed so $\triangle BAF$ is fixed are its components.

Anyway... so

$FG = AG - AF$ And $\frac {AF}{AB} = \frac {AG}{AC} = \frac {AG}{AC + BC}$ so $AG= \frac {AF}{AB}AC = \frac{AF}{AE}(AB + BC)$ so

$FG = \frac{AF}{AB}(AB + BC)- AF=$

$AF(\frac {AB + BC}{AB} - 1)=$

$AF(1 + \frac {BC}{AB} - 1) =$

$AF \frac {BC}{AB}$.

So what are $AF, AB$

$EF = DF\sin \angle D = 24.5 \sin \angle D$.

$BE = 8$ so $BF = 8+24.5 \sin \angle D$.

$AB = DE = 24.5\cos \angle D$.

So $AF = \sqrt{(8+24.5 \sin \angle D)^2 + (24.5\cos \angle D)^2}$

$=\sqrt{64 +2*8*24.5\sin\angle D +24.5^2 \sin^2 \angle D +24.5^2\cos^2 \angle D}=$

$\sqrt{64 + 2*8*24,5\sin\angle D + 24.5^2}$

And so

$FG = AF(\frac {BC}{AB}) =$

$\sqrt{64 + 2*8*24,5\sin\angle D + 24.5^2}\frac {BC}{24.5\cos \angle D}$

.... But you should double check my work.


The relationship between FG and BC is non-trivial, due to all the constrains. Here is a quick way to find it, as well as a trick to solve it.

Consider the right triangle DFE, which allows the equation below, $$DF^2=FE^2+ED^2$$ or, $$ 24.5^2 = \left( 9-x\right)^2 + BC^2 \left( \frac{17}{x} -1 \right)^2 \tag{1}$$

with $x=\sqrt{FG^2 - BC^2}$.

So, given $BC$, solve for $x$ first from (1); and then obtain $FG=\sqrt{x^2+BC^2}$.

  • $\begingroup$ Why the 17/x-1 term? I'm not following that part. $\endgroup$
    – matt
    Aug 13 '19 at 13:39
  • $\begingroup$ $AB^2=BC^2 \left(\frac{17}{x}-1 \right)^2$, which follows from similar triangles, $AB/BC=AF/FG=(17-x)/x$ $\endgroup$
    – Quanto
    Aug 13 '19 at 13:57

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