Progression of a Trig Solution

On my last birthday, one of my more mathematical-minded friends posted the following picture on my Facebook page:

Since I love solving trigonometry problems, I immediately tried to solve it.

First Attempt: Law of Sines

At first glance, I noticed that if I assigned one of the missing angles with a variable, I could use some angle sum properties to label every other angle in terms of that variable.  Letting ∠AKC = t, I found that ∠DKG = 60° – t, ∠ACK = 150° – t, ∠KCD = t + 30°, ∠KDC = 90° – t, and ∠KDG = t + 90°.

I also noticed that since ∠A = ∠G = 30°, △AKG is an isosceles triangle, so I labeled AK = KG = y.

Since I had a lot of angles and sides opposite to each other, I used the law of sines on △AKC and △DKG to obtain two equations with two variables:

which can be solved numerically to t ≈ 7.589 and y ≈ 60.044.

Then since AK cos 30° = ½ AG, by substitution 60.044 cos 30° = ½ (13 + x + 48), and this can be solved to x = 43.

While the answer amusingly worked out to my age, the process involving complicated decimals through numerical methods was less than satisfactory.  I wondered if there was a better way to arrive at the solution.

Second Attempt: Law of Cosines

The next logical attempt after trying the law of sines was to try the law of cosines.  With the knowledge that t works out to a random decimal, I decided to try and avoid that and stick to the integer angles of 30° and 60° instead.

By the law of cosines on △ACK, CK² = AK² + AC² – 2 AK AC cos 30° = y² + 169 – 13√3y.

By the law of cosines on △DGK, DK² = GK² + DG² – 2 GK DG cos 30° = y² + 2304 – 48√3y.

By the law of cosines on △CDK, CD² = CK² + DK² – 2 CK DK cos 60° = CK² + DK² – CK DK.

By substitution on the last equation, x² = y² + 169 – 13√3y + y² + 2304 – 48√3y – √(y² + 169 – 13√3y) √(y² + 2304 – 48√3y).

Also, since AK cos 30° = ½ AG then y cos 30° = ½ (13 + x + 48) or y = (x + 61) / √3.

After substitution, the equation can be (eventually) manipulated to (x – 43)(x + 43)(61x + 1849) = 0, which solves to x = 43 for x > 0.

Although this method was long, it did eliminate the need for numerical methods with complicated decimals, so that was a little better.  I noticed that there were several iterations of the form 13 and 48 throughout this method, so I wondered if a general solution where a = 13 and b = 48 could shed some light onto why the solution worked out to such a nice integer.

Third Attempt: Law of Cosines with General Variables

Using the same steps from before, but with a = 13 and b = 48, then:

By the law of cosines on △ACK, CK² = AK² + AC² – 2 AK AC cos 30° = y² + a² – a√3y.

By the law of cosines on △DGK, DK² = GK² + DG² – 2 GK DG cos 30° = y² + b² – b√3y.

By the law of cosines on △CDK, CD² = CK² + DK² – 2 CK DK cos 60° = CK² + DK² – CK DK.

By substitution on the last equation, x² = y² + a² – a√3y + y² + b² – b√3y – √(y² + a² – a√3y) √(y² + b² – b√3y).

Also, since AK cos 30° = ½ AG then y cos 30° = ½ (a + x + b) or y = (x + a + b) / √3.

After substitution, the equation can be (eventually) manipulated to (x – √(a² + b² – ab))(x + √(a² + b² – ab))((a + b)x + (a² + b² – ab)) = 0, which solves to x = √(a² + b² – ab) for x > 0.

Now I recognized that the general solution x = √(a² + b² – ab) can be re-written as x = √(a² + b² – 2 ab cos 60°), the law of cosines for a 60° triangle!  But why are the segments a, b, and c linear in the diagram, and not part of a 60° triangle?  Can the diagram be manipulated to make solving it easier?

Fourth Attempt: Diagram Manipulation

I noticed that since AK = KG, △ACK can be rotated so that AK matches up to KG:

Then ∠KGC’ = ∠DGK + ∠KGC’ = 30° + 30° = 60°, and ∠DKC’ = ∠DKG + ∠GKC’ = (60° – t) + t = 60°.

Now here’s something interesting: △DKC ≅ △DKC’ by SAS (DK = DK, ∠DKC = ∠DKC’ = 60°, and KC = KC’), so that DC’ = CD = x.

Then by the law of cosines on △DGC’, x = √(DG² + GC’² – 2 DG GC’ cos 60°) = √(48² + 13² – 48 ⋅ 13) = 43.

This method was much easier and much more elegant than the other methods!

Extending the Problem

I’m always looking to find new trig problems to either give to my math students or to post on math forums, especially one with a clever solution like this one.  There are plenty of other 60° triangle integer triples that could have been used, like (3, 8, 7), (5, 8, 7), (7, 15, 13), etc. (see here for more), but I liked the following problem because of its unexpected solution:

This can be manipulated as:

so that by the law of cosines on △DGC’, √2 = √(x² + 1² – 2 ⋅ x ⋅ 1 cos 60°), which for x > 0 solves to x = ½(1 + √5) = φ, the golden ratio!

Similar problems with different angles can also be created.  For example, this problem features an equilateral triangle, where the base angles are 60° and the top middle angle is 30°:

This can be manipulated as follows:

so that by the law of cosines on △DGC’, x = √(3² + 5² – 2 ⋅ 3 ⋅ 5 cos 120°) = 7.

Finally, this problem features an isosceles right triangle, where the base angles and the top middle angle are all 45°:

Its diagram can be manipulated as follows:

so that by the Pythagorean Theorem on △DGC’, x = √(5² + 12²) = 13.  (Interestingly, since this solution only uses triangle congruency and the Pythagorean Theorem, this problem could have been solved by the ancient Greeks!)

Conclusion

There are often different ways to solve a trig problem, with some more difficult than others.  It can be extremely satisfying if you are able to find a clever and easy solution, especially after several attempts at a problem.

SSA Congruence for Obtuse Triangles

In most Geometry classes, students are taught five different ways to prove two triangles congruent: SSS, SAS, ASA, AAS, and HL.  Each “S” stands for “side”, and each “A” stands for “angle”, so SSS means that two triangles can be proven congruent if all three sides are congruent, SAS means that two triangles can be proven congruent if two sides and the included angle are congruent, ASA means that two triangles can be proven congruent if two angles and the included side are congruent, and AAS means that two triangles can be proven congruent if two angles and the non-included side are congruent.  The last congruence, HL, stands for “hypotenuse-leg”, which means that two right angle triangles can be proven congruent if one leg and the hypotenuse are congruent.

Traditional Triangle Congruence Theorems

Students are also taught that SSA (or ASS) is not a way to prove triangles congruent, because in some cases two different triangles can be formed by two given sides and a non-included angle.  With a picture of this counter-example and a predictable lame joke about how we should have no postulates named after a donkey, the teacher (or text) closes the discussion on the matter.

SSA Counter-Example

However, SSA is a legitimate congruence theorem if the given angle is not an acute angle.  This is already evident for right angle triangles, because the HL congruence theorem is really an SSA congruence theorem in disguise.  But it is also true when the given angle is obtuse as well.  In the typical counter-example for SSA, the law of sines and the identity sin(x) = sin(180° – x) can be used to prove that the two different (non-right) triangles have a second angle that is always acute in one case and always obtuse in the other case:

01	by the law of sines, sin B = b/a sin A
02	by a trigonometric identity, sin x = sin (180° – x)
03	therefore, two possible solutions for ∠B are x and 180° – x
04	in a non-right triangle, x is either acute or obtuse
05	if x is acute, then 180° – x is obtuse
06	if x is obtuse, then 180° – x is acute
07	either way, one solution for ∠B is acute and one solution for ∠B is obtuse

However, if the given first angle is already obtuse, then there cannot be a second obtuse angle because the angle sum of the triangle would be more than 180°.  This leaves only one possible way to form the triangle, which means SSA is a legitimate congruence theorem when the given angle is obtuse.

SSA Congruence

(if the given angle is obtuse)

                                                                                  

Most mainstream Geometry texts only teach five different ways to prove two triangles congruent (SSS, SAS, ASA, AAS, and HL), and only mention SSA as an incorrect way to prove two triangles congruent.  Indeed, SSA is incorrect if the given angle is acute, because the given criteria can produce two different possible triangles.  However, if the given angle is right or obtuse, only one triangle can be produced, in which case SSA is a legitimate congruence theorem.