Can you find the angle X? | (Justify your answer) |

Learn how to find the value of angle X in the triangle. Important Geometry and Algebra skills are also explained: Trigonometry; Exterior angle theorem; equilateral triangles; isosceles triangles. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Can you find the angle...
Can you find the angle X? | (Justify your answer) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
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[@SkinnerRobot]

You make it look so easy. Thank you, PreMath.

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@engralsaffar]

Angle CEB=135 therfore angle B=30 degrees. Then, connect a line from D to O perpendicular to CB, and draw a cemicircle to with D, O and B at its circumference, from the circle theorem this makes DE, BE and OE radii in the circle. Angle ODE=60 degrees (30-60-90), and angle DOE is also 60, therefore triangle DOE is an equilateral triangle, and angle OED=60. Therefore angle OEC=15, and OCE is an isosceles triangle with OC=OE, and it is also equal to OD, therefore ODC is also an isosceles triangle with a 90 angle, Therefore the bases are equal to 45 degrees, therefore x+15=45 x=30 degrees

[@jimlocke9320]

As I noted yesterday, the 15°-75°-90° triangle, while not considered "special" in geometry, appears quite frequently in problems. Since we are allowed to use our notes when solving these problems, I recommend that everyone add the properties of the 15°-75°-90° triangle to their notes (or just memorize the properties)! Ratio of sides short:long:hypotenuse (√3 - 1):(√3 + 1):2√2 Ratio short/long = (√3 - 1)/(√3 + 1) = (2 - √3)/1 Ratio long/short = (√3 + 1)/(√3 - 1) = (2 + √3)/1 In this problem, we see AD having length (2 - √3) and AC length 1, so the ratio AD/AC = (2 - √3)/1 and ΔDAC is a 15°-75°-90° right triangle. The short side is opposite the 15° angle.

[@uwelinzbauer3973]

Hallo Professor, I exactly used the same approach as shown in the video. Thanks for the interesting geometric puzzle! I wish a happy 😊 weekend to you and the channel visitors!

[@quigonkenny]

As ∠DEC is an exterior angle to ∆CEB at E, ∠DEC = ∠BCE+∠EBC. ∠DEC = ∠BCE + ∠EBC 45° = 15° + ∠EBC ∠EBC = 45° - 15° = 30° Draw DF, where F is the point on BC where DF and BC are perpendicular. As ∠DBF = 30° and ∠BFD = 90°, ∠FDB must equal 60° and ∆BFD is a 30-60-90 special right triangle. If DF = a, then DB = 2a and BF = √3a. As DB = 2a, DE = EB = a. Draw EF. As DF = DE = a and ∠FDE = 60°, then ∆FDE is an equilateral triangle, ∠DEF = ∠EFD = 60°, and EF = DF = DE = a. As ∠DEF = 60° and ∠DEC = 45°, ∠CEF = 60°-45° = 15°. As ∠FCE = 15° as well, ∆EFC is an isosceles triangle and FC = EF = a. As FC = DF = a and ∠DFC = 90°, then ∆DFC is an isosceles right triangle, and ∠CDF = ∠FCD = (180°-90°)/2 = 45°. As ∠FCD = 45° and ∠FCE = 15°, ∠ECD = x = 45°-15° = 30°.

[@NIKHIL-ko5xm]

You really did this on your own ?! 😮

[@magamoodley3975]

Use midpoint and parallel lines Mid point theorem. Try neat solution

[@alexundre8745]

Bom dia Mestre Obrigado pela aula

[@CliffordMorris-ls9lc]

Basic angles if a triangle equals 180 was all you needed to see ACB was 60degrees and ACE was 45 degrees

[@soli9mana-soli4953]

since this problem is substantially equivalent to the one previously posted, without doing any calculations I would say that x = 30°

[@NinaDolgan]

This supplement has made my nervous system function more steadily.

[@黃羅賓]

It has a easier way to solve this problem, make a line from E to CB

[@alexj8940]

Make point O on AB such as EO=EB. Angle EOB=EBO=30. Therefore angle OEB=30. Therefore OC=EO=DE. Connect O with D, we’ve got equilateral triangle DOE. therefore DOB=90. DO=CO and COD=90. DO=CO, DCO=45 therefore X=45-15=30

[@fantasticyt9951]

As DE=EB Considering triangle BCD, We can say that CE is the median of triangle BCD Hence according to triangle laws, x=2 x 15 = 30 degrees

[@prossvay8744]

x=30° (Law of the sine).❤❤❤

[@marcelowanderleycorreia8876]

Drop a perpendicular line from point D to line BC - DP lenght. Lenght CP = a, since CPB is a special 30 - 90 - 60, PD is also = a, hence x + 15 must be iqual to 45 degrees, x = 45 - 15 = 30 degrees. I spent all day to view this...

[@PrithwirajSen-nj6qq]

BE=DE= m AD=n AB= 2m +n AC =AE = m +n (as 🔺 ACE is 90-45-45 triangle ) Ang ADC =45+x tan (45+x)=AC /AD =(m+n)/n=m/n. +1 Ang CBD=45-15=30 degs tan 30=AC/AB= 1/√3 = (m+n)/(2m+n) m√3 -2m= n-n√3 m(√3-2)=n(1-√3) m/n=(1-√3)/(√3-2) Hence tan (45+x) =[(1-√3)/(√3-2)] +1 = 2+√3= tan 75 deg Then x =75-45=30 degs Note -- angles measured in degrees in all cases

[@cyruschang1904]

The big triangle is a 30° - 60° - 90° triangle with the base (y + x + x) and height (x + y) 2x + y = (x + y)✓3 (2 - ✓3)x = (✓3 - 1)y x = (✓3 - 1)y/(2 - ✓3) = (✓3 + 1)y (x + y)/y = ✓3 + 2 y/(x + y) = 2 - ✓3 x = 60° - 15° - arctan(2 - ✓3) = 45° - arctan(2 - ✓3)

[@allanflippin2453]

The solution is easy if we know that tan 15 = 2-sqrt(3). But if we don't know that, how can it be derived?

[@michaeldoerr5810]

The answer x = 30°. I think that this application is very similar to one of Math Booster favorite constructions. And furthermore I think that this easy enough process is ONLY applicable to 30°-60°-90° triangles.

[@peta1001]

How did you confirm/prove (in minute 6:00) that segment DE is equal to segment EB?

[@luisfilipe2023]

It’s given by the problem

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال DB=1 AD=a ACD=y x+y=45 ...... tany=tan15 y=15 x=30

[@imetroangola17]

*Solução:* Seja DE=BE=a. Além disso, o ângulo B= 45° - 15°= 30°. Aplicando a Lei dos senos no ∆CED: sen x/DE = sen 45°/CD *sen x/a = sen 45°/CD (1)* Aplicando a Lei dos senos no ∆CDB: DB/sen (x + 15°) = CD/sen B *2a/sen (x + 15°) = CD/sen 30° (2)* Multiplica (1) e (2): 2 sen x/sen (x + 15°) = sen 45°/sen 30° sen x/sen (x + 15°) = (sen 45°/sen 30°)×1/2 sen x/sen (x + 15°)=(sen 45°/sen 30°)×sen 30° sen x/sen (x + 15°) = sen 45° sen x/sen (x + 15°) = 1/√2 √2 sen x = sen (x + 15°) Facilmente, *x=30°* satisfaz está última expressão.

[@marcgriselhubert3915]

Let's use an orthonormal center A and first axis (AB). We have A(0; 0) E(c; 0) C(0; c) with c = AE =AC The equation of (CB) is y - c = -tan(30°).x or y = (-sqrt(3)/3).x. The intersection of (CB) with (AB) is then B((sqrt(3).c; 0) and EB = (sqrt(3) -1).c As EB = DE, we then have DE = (sqrt(3) -1).c and D((2 -sqrt(3)).c; 0) Now we have VectorCD((2 -sqrt(3)).c; -c) and CD^2 = (c^2).(4+3-4.sqrt(3) +1) = (c^2).(8 -4.sqrt(3)) and CD = c.sqrt(2).(sqrt(3) -1) Also we have VectorCE(c; -c) and CE^2 = 2.(c^2) and CE = c.sqrt(2). In triangle CDE we have DE^2 = CD^2 + CE^2 - 2.CD.CE.cos(x), so: (c^2).(3+1-2.sqrt(3)) = (c^2).(8 -4.sqrt(3)) + 2.(c^2) -2.c.sqrt(2).(sqrt(3) -1).sqrt(2).c.cos(x) We simplify: 4.(sqrt(3) -1).cos(x) = 6 -2.sqrt(3) or cos(x) = (3 -sqrt(3))/(2.(sqrt(3) -1)) = sqrt(3)/2, which gives that x = 30° (In fact the length c has no effect, we could have choosen c = 1 at the beginning without changing the generality of the problem. This problem is quite similar to the one from yesterday)

[@giuseppemalaguti435]

t=CE,a=DE..t/sin(135-x)=a/sinx...t/sin30=a/sin15...divido rimane ctgx=(1/sin15-√2)/√2=1/(√2sin15)-1=√3...x=30

[@nenetstree914]

30

[@unknownidentity2846]

Let's find x: . .. ... .... ..... First of all we calculate some angles: ∠ACE = 180° − ∠CAE − ∠AEC = 180° − 90° − 45° = 45° ∠ACB = ∠ACE + ∠BCE = 45° + 15° = 60° Since ABC and ACE are right triangles, we can conclude: tan(∠ACB) = AB/AC ⇒ AB = tan(∠ACB)*AC = tan(60°)*AC = √3*AC tan(∠ACE) = AE/AC ⇒ AE = tan(∠ACE)*AC = tan(45°)*AC = AC With DE=BE we obtain: AB = AD + DE + BE = AD + 2*DE = √3*AC AE = AD + DE = AC ⇒ AD = 2*AD − AD = 2*AD + 2*DE − AD − 2*DE = 2*(AD + DE) − (AD + 2*DE) = 2*AE − AB = 2*AC − √3*AC = (2 − √3)*AC Now we are able to calculate the value of x: tan(x) = tan(∠DCE) = tan(∠ACE − ∠ACD) = tan(45° − ∠ACD) = [tan(45°) − tan(∠ACD)]/[1 + tan(45°)*tan(∠ACD)] = (1 − AD/AC)/(1 + AD/AC) tan(x) = [1 − (2 − √3)]/[1 + (2 − √3)] = (1 − 2 + √3)/(1 + 2 − √3) = (√3 − 1)/(3 − √3) = (√3 − 1)/[√3*(√3 − 1)] = 1/√3 = (1/2)/(√3/2) tan(x) = sin(x)/cos(x) = sin(30°)/cos(30°) ⇒ x = 30° Best regards from Germany

[@gnanadesikansenthilnathan6750]

X=30 degrees

[@murdock5537]

x = 30°

[@左近允庸孝]

昨日見ましたね！☺️

[@PreMath]

フィードバックありがとうございます❤️

[@marcgriselhubert3915]

A remark: In your method somewhere you have tan(theta) = 2 - sqrt(3), it is good to know that theta = Pi/12 rad = 15° For theta = Pi/12 rad = 15°, then sin = (sqrt(6) - sqrt(2))/4; cos = (sqrt(6) + sqrt(2))/4; tan = 2 -sqrt(3); cotan = 2 + sqrt(3) Also interesting: for theta = Pi/8 rad = 22.5°, then sin = (sqrt2 -sqrt(2))/2; cos = (sqrt(2 +sqrt(2))/2; tan = sqrt(2) - 1; cotan = sqrt(2) + 1.

[@JSSTyger]

30

[@ponnananthik5597]

30