Can you find Angle X? Justify your answer! | Quick & Simple Explanation

#OlympiadMathematics #OlympiadPreparation #CollegeEntranceExam

[fevengr]

I’m retired and enjoy trying your problems so that I can preserve whatever brain cells are left. For this problem, I discovered what was most likely the most difficult solution. I created equations for line segments AC and CB with point C located at coordinate (0, 0). Then I drew a line perpendicular to AB that intersected point C. The point where it intersected AB I labeled as point Z. To simplify the problem as much as possible I set the length for segment DZ as 1 unit. Without going into horrendous detail, I used the line equations to determine that segment CZ was also 1 unit long thus making the angle 45 degrees. Thanks for your challenging videos!

[MrMichelX]

Did not think to this kind of point of view at all...! Thanks !

[davidfromstow]

Your explanations always make to so obvious (afterwards!). I get cross with myself for so rarely spotting the solution! Thank you.

[d-hat-vr]

This is a very cool question and the added "think outside the box" construction lines are clever! Reminds us to consider completing a 30-60-90 right triangle when we see a length 2k side adjacent to a 30° angle.

But I wanted to see if I could find an answer by "thinking inside the box" instead. (Reminder: I'm 53 years old, have no natural talent for math or logic, and my brain is very slow). Here's the answer I came up with:

1. For ease of convenience, label ∠BCD as ∠y and label ∠ACD as ∠z. Also, let's call the length AD=DB=a.

2. By Law of Sines we find sin y/a = sin 30°/DC and sin z/a = sin 15°/DC.

3. Combining these equations produces sin y/sin z = sin 30°/sin 15°. From this we might guess y=30° and z=15° but that's not the correct answer since we know y+z=135°

4. Applying the double angle for sine, we can rewrite the equation as sin y/sin z = 2⋅cos 15° (Note that cos 15° = cos(45° - 30° ) so not too hard to compute with cosine angle difference formula.)

5. Recalling that y+z=135°, the equation can be further rewritten as sin (135°-z)/sin z = 2⋅cos 15°. Applying sine difference formula to the denominator gives (sin 135°⋅cos z - cos 135°⋅sin z)/sin z = 2⋅cos 15°.

6. Now sin 135° = 1/√2 and cos 135° = -1/√2, so doing some algebra on the equation and multiplying both sides by √2 we find (cos z + sin z)/sin z = 2⋅cos 15°⋅√2

7. Separating the left side of the equation and calculating out the right side of the equation gives cot z + 1 = √3 + 1, hence z = arccot(√3) = 30°. From this we can find y=105° and then x=45°.

Note that solving for z is easier than solving for y, since the expression one ends up with at the end of an analogous calculation does not obviously resolve to y=105°, unlike arccot(√3) that is more easily seen to be 30°. The corollary that sin 105° ⋅ sin 15° = (sin 30°)² seems surprising at least to me.

[josephmcneil]

I’ve watched all the way through and absolutely love the way you got to the answer through higher level/simple angles and properties.
With that said, 4:30 is where I’d have pulled a ratio of distance and run a trig function to solve.
I’be downloaded this to show my youngest when he gets to geometry. Great job!

[Ramkabharosa]

We can let |AD| = |DB| = 1 unit. Also let |CD| = b units, angle ACD = y, & angle BCD = z. Then y + 15° = x, so y = x - 15° and sin(z) = sin(180° - 30° - x) = sin(x+30°). Now by the Sine Rule, sin(x-15°)/1 = sin(15°)/b & sin(x+30°)/1 = sin(30°)/b. So from the 2nd equation, 1/b = sin(x+30°)/sin(30°). Substituting in the 1st eq., we get sin(x-15°) = sin(15°).sin(x+30°)/ sin(30°). So sin(30°).sin(x-15°) = sin(15°).sin(x+30°). Expanding both sides we get,

sin(30°). [sin(x).cos(15°) - sin(15°).cos(x) ] = sin(15°). [sin(x).cos(30°) + cos(x).sin(30°)]. So [sin(30°).cos(15°) - sin(15°).cos(30°)].sin(x) = [sin(30°).sin(15°) + sin(30°).sin(15°)].cos(x). Thus sin(x)/cos(x) = 2.sin(30°).sin(15°)/sin (30°-15°) = 2.(1/2).sin(15°)/sin (15°) = 1. So tan(x)=1 & hence x=45°. In case, they're needed: y= x-15°= 30°, z= 180°-30°-x =105°, & b= sin(15°)/sin(x-15°) = sin(15°)/sin(30°) = sin(15°)/(1/2) = (√6 - √2)/2. Trig. is king!
.

[lalaromero]

Looked easy at first then realized it was a nightmare to solve 😂

[wes]

Simple solution using law of sines. Let BD = y and AD = BD = z. Angle BCD = 150-x and angle ACD = x-15. Thus
sin15/y = sin(150-x)/z and sin30/y = sin(150-x)/z. Eliminate y and z to obtain
tanx = and x=45°.

[vidyapatechawan]

Very nice.... Again recollected school/college time... Geometry.. Thanks sir

[AnonimityAssured]

I do like this problem, and I love your solution. There is a sort of shortcut that makes use of the interaction between a square and an equilateral triangle external to it and sharing a side with it. Obviously, the internal angles of the triangle are each 60°. Less obviously, if T is the vertex of the triangle not touching the square, and line segments are drawn from T to each of the two corners farthest from it, then those line segments subtend an angle of 30°.

[trendingNOW]

I love these as I have never learned any of this at school. My first action was to create a regular quadrilateral starting at point C to include point D as the opposite corner. This gave an angle of 45. I then watched the video to see how to do it correctly. )))

[luigipirandello]

Very nice solution. Easy to understand. Thank you Sir.

[fastwalker]

“… and here’s our much nicer looking diagram!”

I love it👍

[vhm]

Nice solution 👍 I have another one that is the same approach as yours.

I doubled the two angles ∠A and ∠B, so that I make a right triangle △ABF, whose ∠A = 30° and ∠B = 60°.
Next we have 2 congruent triangles *△BCD ≅ △BCF* (because of BD = BF; ∠CBD = ∠CBF = 30° and they have the common side BC).
The point C has to be the center of the inscribed circle of *△ABF* (intersection of two bisectors AC and BC), then we have ∠BFC = 90°/2 = 45°.
Finally the value of x is equal to ∠BFC = 45°

[erdalyuksel]

I wait for new problem which is more difficult than one. Thaks for all

[robertlynch]

OK, your derivation is definitely BEAUTIFUL geometry! I however found it a trigonometric way.

Let
   𝒉 = height of triangle.
   𝒎 = length of one of the two congruent segments.
   (𝒎 + 𝒂) is segment to left
   (𝒎 - 𝒂) is segment to right

Seeing that the right triangle is a 30–60–90, then
   𝒉 = (𝒎 - 𝒂)/√3
   𝒉 = (𝒎 - 𝒂)*√⅓

The tangent of 15° = 0.267949 in general. This I will call T₁₅ for simplicity.
   T₁₅(𝒎 + 𝒂) = (𝒎 - 𝒂)√⅓

Léts expand and flip the 𝒎's and 𝒂's around to isolate
   T₁₅𝒎 + T₁₅𝒂 = √⅓𝒎 - √⅓𝒂 … rearrange
   T₁₅𝒂 + √⅓𝒂 = √⅓𝒎 - T₁₅𝒎 … combines to
   (T₁₅ + √⅓)𝒂 = (√⅓ - T₁₅)𝒎

Now, arbitrarily set (𝒎 = 1), in order to solve 𝒂
   𝒂 = (√⅓ - T₁₅) / (√⅓ + T₁₅)
   𝒂 = (0.577350 - 0.267949) / (0.577350 ⊕ 0.267949)
   𝒂 = 0.366025

Armed with that, now figure height of triangle
   𝒉 = √(⅓)(𝒎 - 𝒂)
   𝒉 = 0.577350 (1 - 0.267949)
   𝒉 = 0.366025

Well, that's the same as 𝒂, so the inscribed △ has ( height = base ), and thus must be a 45–45–90.

   α = 45°

Tada.

⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅

[Mete_Han]

Another great question from PreMath, thank you.

[zplusacademy]

Wow 😳 sir ❤️🙏🙏🙏🙏 superb creation 🙏🙏🙏❤️

[HappyFamilyOnline]

Very nice explanation👍
Thanks for sharing🎉

[antibulling]

angle en CAB on a les deux angles de la base du triangle abc et la somme des angles est de 180° angle y cab = 180 - 15 - 30° = 180 - 45 = 135° comme AD = DB l'angle CDB = CAB / 2 = 135/2 et x + 135°/2 + 30° = 180° => x =