# Problem Statement

An acute triangle intersects with another triangle to form a four-sided polygon. Given all angles, and given the distance of two sides of the polygon, find the other two sides of the polygon. See image (not drawn to scale).

• The angles that form the acute triangle are: 11°, 81°, 88°
• The angles that form the other triangle are: 71°, 81°, 28°
• side a = 57 inches
• side b = 70.25 inches
• find side c & d

# Attempt

The acute triangle is actually an isosceles triangle (11°, 84.5°, 84.5°) with an arbitrary line drawn through it. I have used the Law of Sines and SOH CAH TOA to find the midpoints of the acute/isosceles triangle. I thought I could use the same techniques to find sides c & d, but I can't seem to figure out how to create an equation that relates the two intersecting triangles. My gut tells me that a calculation should be possible given the constraining angles and distances.

# Background

I am building a tree-house, where the the four-sided polygon represents the main deck. I need to figure out the correct spacing of joists that will lay between the legs of the isosceles triangle. Because I am working in a live tree, nothing is square and true. I expect I will need to install 6 joists, but I won't know for sure unless I lay it out. Since it is such a pain in the neck to put even one joist down, I want to plan ahead. I'd like to minimize the amount of time tying into my safety line, going up and down, measuring for length, measuring for angle, and test fitting, and so forth.

I know if I carefully sketch it out, mechanical-drawing style, I could measure the distance on the paper, but that has its own challenges. I've also attempted to sketch it out using computer software, which for me has a high learning curve. Using a calculation allows me to keep squishy things in check, like material inconsistencies, flexing of fasteners, and movement of the tree itself. It also gives me a better way to check my work. Lastly, without having to use my Dad Voice, I want my kids to understand that there is more to math than just homework.

• It's not clear which lines are a, b, c, d. From your description, the overall shape of the polygon is like a concave kite or arrowhead. Are a, b,c ,d the sides of this shape or of the original triangles? Apr 22 '21 at 4:18
• Sorry, I think I've totally misinterpreted your diagram. I think you mean the polygon that is kite-shaped with internal angles 71, 81, 88 and 120 degrees, and "bottom" side = 70.25, "top" side = 57. Right? Apr 22 '21 at 4:59
• @robert-timmer-arends, your second comment is correct. The polygon has internal angles of 71°, 81°, 88° and 120°. The "top" side is 57 inches and the "bottom" side is 70 1/4 inches. Apr 22 '21 at 20:09

Starting with the calculation of other angles in the relevant triangle gives

Then we can apply the sine rule twice: $$\frac{70.25}{sin11} = \frac{c+u}{sin88} = \frac{d+v}{sin81}$$ $$\frac{57}{sin11} = \frac{u}{sin60} = \frac{v}{sin109}$$ The first gives $$c+u=367.94$$ and $$d+v=363.64$$

The second gives $$u=258.71$$ and $$v=282.45$$

From these $$c=109.23$$ and $$d=81.19$$

• Ah, this method I was trying to use. But I couldn't figure out how to get the two equations to work together. Thanks for hitting the mark, and showing me how to do it correctly. Apr 23 '21 at 2:16

This figure ...

... tells us

\begin{align} b \sin(P+R) &= c\sin R+a\sin(P+Q+R) \\[4pt] \to\quad c &= \frac{-a\sin(P+Q+R)+b\sin(P+R)}{\sin R} \tag{1} \end{align}

Another dissection (left as an exercise to your kids) tells us, $$d =\frac{- a \sin(P + Q) + b \sin P}{\sin R} \tag{2}$$ With $$a=57$$, $$b=70.25$$, $$P=81^\circ$$, $$Q=28^\circ$$, $$R=11^\circ$$, these become

$$c = 109.239 \qquad d = 81.1835 \tag{\star}$$

A GeoGebra sketch validates these measurements:

Enjoy the treehouse ... and the math lesson! :)

(Come to think of math lessons: It may be instructive to check the Edit History for my first pass at this problem. Seeing the final forms for $$c$$ and $$d$$ is what suggested to me that a more-direct derivation was possible.)

• I never would have thought of this solution, but I like the more direct derivation (and the visuals). Apr 23 '21 at 2:27

Draw the diagonals in the quadrilateral

and use the cosine theorem: $$\begin{cases}\color{blue}{f^2}=c^2+70.25^2-2c\cdot 70.25\cos81^\circ=57^2+d^2-2d\cdot 57\cos120^\circ\\ \color{green}{e^2}=c^2+57^2-2c\cdot57\cos71^\circ=d^2+70.25^2-2d\cdot70.25\cos88^\circ\end{cases}$$ Wolfram answer: $$c\approx 109.24, d\approx 81.18$$.

• I dabbled a bit with cosine theorem, but I guess I should have given it some more effort. Well done. Apr 23 '21 at 2:19
• See the addition for the details! Apr 24 '21 at 6:10