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.