Can you solve for the angle?

0:00 problem
1:17 solution 1
7:28 solution 2

Math StackExchange

Reddit

Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!

If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.

If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.

Book ratings are from January 2023.

My Books (worldwide links)

My Books (US links)
Mind Your Decisions: Five Book Compilation
A collection of 5 books:
"The Joy of Game Theory" rated 4.3/5 stars on 290 reviews
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 4.1/5 stars on 33 reviews
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.2/5 stars on 54 reviews
"The Best Mental Math Tricks" rated 4.3/5 stars on 116 reviews
"Multiply Numbers By Drawing Lines" rated 4.4/5 stars on 37 reviews

Mind Your Puzzles: Collection Of Volumes 1 To 3
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 112 reviews
"Math Puzzles Volume 2" rated 4.2/5 stars on 33 reviews
"Math Puzzles Volume 3" rated 4.2/5 stars on 29 reviews

2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.

My Blog

Twitter

Instagram

Merch

Patreon

Press

[jimmytaco]

“…this is equal to 0.577, which is easily recognizable as 1 divided by the square root of 3”

[jimlocke]

As others have pointed out, the equation tan(x) = (tan(50°)tan(10°))/tan(20°) at 10:10 can be solved for an exact value of x without using numerical values. The first step is to replace 1/(tan(20°) with the exact equivalent tan(70°). Then we recognize that 70° = 60° + 10° and 50° = 60° - 10°, so tan(x) = tan(60° - 10°)tan(10°)tan(60° + 10°). Now, we can use the identity tan(3Θ) = tan(60° - Θ)tan(Θ)tan(60° + Θ). Let Θ = 10° and we find that x = 30°.

To prove the identity for Θ = 10°, we use the tangent sum of angles and difference of angles formulas. So, tan(a + b) = (tan(a) + tan(b))/(1 - tan(a)tan(b)) and tan(a - b) = (tan(a) - tan(b))/(1 + tan(a)tan(b)). We substitute a = 60°, so tan(60°) = √3, and b = 10°. So, after simplification, tan(60° - 10°)tan(60° + 10°) = (3 - tan²(10°))/(1 - 3 tan²(10°)). For the left side, we first compute tan(20°) = tan(10° +10°) by either the sum of angles or double angle formula, finding tan(20°) = (2tan(10°))/(1 - tan²(10°)). Then we compute tan(30°) = tan(20° +10°). Simplifying and factoring, we find tan(30°) = tan(60° - 10°)tan(10°)tan(60° + 10°).

In the real world, all measurements have tolerances. In geometry, measurements often have exact values and the problem solution may require that the measurements are exact. For example, a triangle is not equilateral unless all 3 interior angles are exactly 60° and a triangle is not a right triangle unless one interior angle measures exactly 90°.

[DeisonPreve]

For the geometrical solution, the problem ended in 5:45. It is suffice to see that x+y=50 from the start, 2x+y=80 (from the isosceles triangule), therefore, x=30.

[Bob]

As an engineer, I would draw the triangles according to specifications, and then measure the wanted angle.

[gobleturky]

Here's a way to finish the trig method without a calculator:

tan10 * tan50 / tan20
= tan10 * tan50 * tan70
= tan10 * tan(60-10) * tan(60+10)
= tan10((tan60 - tan10) / (1 + tan60tan10))((tan60 + tan10) / (1 - tan60tan10))
Since tan60 = sqrt(3):
this equals tan10 * (3 - tan^2(10))/(1 - 3tan^2(10))
= (3tan10 - tan^3(10) / (1 - 3tan^2(10))

Now, from the somewhat obscure triple-angle formula tan(3x) = (3tanx - tan^3(x)) / (1 - 3tan^2(x)),
This equals tan(3 * 10) = tan30. So x = 30.


*the triple-angle thing can be derived by tan(2x + x), or alternatively set z = 1 + bi on the complex plane (so b = arg(z)) and compute the argument of z^3.

[BigAsciiHappyStar]

I found an alternative solution:

Extend BA to Q such that triangles BDA, CQA are similar. Then triangles BCA, DQA are also similar (side-angle-side). Angle chasing yields DQ bisects angle BQC. Hence D is the incentre of triangle BQC. Hence x = 1/2 * 60 = 30. QED.

[WillRennar]

My thought process once you established isosceles triangle BCD": "Angles D"BC and D"CB are both 50°, so angle BD"C msut be 80°. Therefore, D'D"C is (80 - 60) 20°, and since it's the vertex of an isosceles triangle, the other two angles must be ((180 - 20)/2) 80°. Angle D"CD' is equal to (50+x)°. 50°+x° = 80°, therefore x=30°.

[markwallen]

I employed a much simpler trig. method? The internal angle at D, within the smaller triangle, appears directly proportional to D's position on AE, compared to AC. As AE is bisected from point C, the internal angle at A( 40* ), increases by .5 times at D( 60* ). 180* - ( 60* + 90* ) = 30 degrees for x ! Who needs a calculator ? 😃

[IvanToshkov]

Correction. Trigonometry + a calculator is overpowered.

[gamefacierglitches]

Solved it the second way in my head, but a tad bit different.

(After getting the 50 degrees)
BE = y
EC = z

y * tan(10) = z * tan(x)
y * tan(20) = z * tan(50)

Divide the equations:
tan(10) / tan(20) = tan(x) / tan(50)

(From here it was the same)

[bobzarnke]

More generally, if the bisected angle is 2θ, the top angle is φ and the desired angle is α, then it's easy to show that tan(α) = tan(θ)/(tan(2θ)tan(φ)). For this problem, this gives the
interesting identity for angles in arithmetic progression: tan(10)=tan(20) tan(30) tan(40).

[nedmerrill]

Very good. I tried it just looking at the angles specified in the figure, but I ended up with 5 equations in 5 unknowns, but the resulting matrix was singular, and then i gave up.
I appreciate the trig solution more. It requires less imagination, which I appreciate.
I admire the geometric construction and the imagination it took.

[TheEulerID]

I got the answer 30 degrees, but had to use trigonometry.

I just set BE=1. Then I generated trig equations

DE = tan(10°)
AE = tan(20°)
EC = AE * tan(40°) = tan(20°)tan(40°)
tan(x) = DE/EC = tan(10°)/(tan(20°)tan(40°))

I used a calculator, which seems like cheating, but no doubt it can be done using trigonometric identities.

Ok. just looked at the geometry answer. I could have looked at that for weeks, and not got it. Thank heavens for trig.

[Z-eng]

Ok, I see many here who also tried solving for an exact value of tan(x) like me and used the complex looking tan formulas to do so, as for me I'll just show how I solved it using only sines and cosines instead, which only makes the formulas easier, but the equations somewhat long, here goes.
So first I turned every tan to its sine and cosine forms, sin(x)*sin(20) / (cos(x)*cos(20)) = sin10*sin50 / (cos10*cos50), then cross multiply, sin(x)*sin20 * cos10*cos50 = cos10*cos50 * sin10*sin50, then I use the product to sum formulae to turn each pair multiplied into a sum of cosines, sina*sinb = 1/2 (cos(a-b) - cos(a+b)) & cosa*cosb = 1/2 (cos(a-b) + cos(a+b)) [note that for cosines a-b is the same as b-a since they're even functions], now substituting and multiplying both sides by 1/4 I get, [cos(x-20) - cos(x+20)]*[cos(50-10) + cos(50+10)] = [cos(x-20) + cos(x+20)]*[cos40 - cos60], multiplying all left terms and all right terms we get, cos40*cos(x-20) + cos60*cos(x-20) - cos40*cos(x+20) - cos60*cos(x+20) = cos40*cos(x-20) - cos60*cos(x-20) + cos40*cos(x+20) - cos60*cos(x+20), we subtract the first and last term on each side from the equation then move the terms a bit to get, 2cos60*cos(x-20) = 2cos40*cos(x+20), we divide by 2 and write the cos60 as its value 1/2, 1/2 * cos(x-20) = cos40*cos(x+20), then do product to sum formula to the right side to get, 1/2 * cos(x-20) = 1/2 [cos(x+20-40) + cos(x+20+40)], multipy equation by 2, and simplify, cos(x-20) = cos(x-20) + cos(x+60), we subtract cos(x-20) from both sides to get, cos(x+60) = 0, arccos(0) is 90deg +360 {we'll ignore the +360 since x is acute}, x+60 = 90, hence x= 30

[ericzhu]

Just a tip, remember that in math olympiad contest (i assume Polish mathematical olympiad uses the same rule as the international olympiad) calculators are not allowed, therefore you should have shown a way to finish the trigonometric way without calculator...

[zey]

@5:40, when you prove that line CD'' is congruent to lines D'D'' and BD'', you can solve for angle D'D''C which must be 50+50+60+x=180 so that angle is 20 degrees.
Then you use the fact that CD'D'' is an isosceles triangle and that the total angle D'CD'' must be (180-20)/2, which is 80 degrees.
Also, let's call angle DCD'' y, for easier equation.
You get an equation pair:
2*x+y=80
x+y=50 (got that earlier, from the starting diagram)
From these you get x=30 (also y=20...)
No need to prove that CDD' is an equilateral triangle or circle from point D'' or arcs of it

[Viktor]

I solved it, too. I used following method to get an analytic solution: I got pretty much the same til the eq. tanx=tan10°*tan50°/tan20°. Then I've written tan10°=tan(30°-10°) and tan50°=tan(30°+20°) in hope to eliminate them. I got It looks scary but after substituting tan30°=1/sqrt3 I got exactly the formula of the reciproc of tan(3*20). As 1/tan60°=tan30°, x=30°. Maybe this looks confusing here in a comment but it was just 6 rows on a piece of paper. I have to admit I had the wikipedia page open for the common trigonometric equations.

[AFSMG]

The trigonometric solution is very clear. Congratulations.

[KlokJammer]

How I did it:

X = arctan(DE/CE)

Because BD bisects ABE, therefore DE is half of AE.
tan(40) = CE / AE --> CE = tan(40)*AE
substituting them in our equation we get
X = arctan[(AE/2) / (tan(40)*AE)] AE cancels out
X = arctan[(1/2) / (tan(40))] move the numerator's denominator to the denominator
X = arctan(1 / 2tan(40))
I had to use the calculator for this but the answer comes to around 30 degrees with rounding errors.

[saetainlatin]

You can find exercises like this in a peruvian book called "Construcciones en Triángulos - Técnicas y Criterios para realizar Trazos Auxiliares" by José Luis Meza Bárcena Cuzcano"