Very nice problem.

@PreMath

12 күн бұрын

Many many thanks, dear 🌹

Thank you! Your method is much more better than mine! 1/ Drop the height CH = h to the base AB and label HD=a We have h=a.sqrt 3 --> h=HB=a+1 so a.sqrt3= a+1-> a=(sqrt3+1)/2-> h=(3+sqrt3)/3 (1) 2/ AH= 2-a=(3-sqrt3)/3 (2) 3/ tan (x) = h/AH= 2+sqrt3 --> x= 75 degrees 😅

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

Thank you!

I love that you can solve the problem just by logic reasoning and without trigonometry!

@PreMath

11 күн бұрын

Excellent! Glad to hear that! Thanks for the feedback ❤️

Method not using construction but a calculator for fast solution in exam: 1. In triangle BCD, angle BCD = 60 - 45 = 15 2. In triangle BCD, find CD by sine rule: CD = (1)(sin45)/(sin15) = 2.732.... 3. In triangle ACD, find AC by cosine rule: AC^2 = CD^2 + 2^2 - (2)(2)(CD)cos60 (Use ANS key to directly input CD from last step.) AC^2 = 6 4. In triangle ACD, by sine rule: sinX = (2.732)(sin60)/sqrt 6 = 0.966 Hence X = arcsin 0.966 = 74.996

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

Elegant Solution

Sine formula to calculate length of CD in triangle CBD. Then cosine formula to determine CA in triangle CAD. Then sine formula to determine x in Triangle CAD.

Nice! I went straight to trigonometry and a combination of the laws of sines and cosines to reach the same result.

Interesting

@user-in4st5nk7g

11 күн бұрын

Very interesting

@PreMath

11 күн бұрын

Excellent! Glad to hear that! Thanks for the feedback ❤️

@user-in4st5nk7g

11 күн бұрын

@@PreMath Thank you!

LBCF=60-45=15° LACD=180°-(60°+x) Let CD=a In ∆BCD 1/sin(15°)=a/sin(45°) so a=sin(45°)/sin(15°) (1) In ∆ACD a/sin(x)=2/sin(180°-(60°+x) a=2sin(x)/sin(60°+x) (2) (1) and (2) sin(45°)/sin(15°)=2sin(x)/sin(60°+x) so sin(45°)sin(60°+x)=2sin(x)sin(15°) sin(45°)=1/√2 sin(60°+x)=sin(60°)cos(x)+sin(x)cos(60°}=√3/2cos(x)+1/2sin(x) 1/√2(1/2)(√3cos(x)+sin(x))=2sin(x)sin(15°) 1/2√2(√3cos(x)+sin(x))=2sin(x)sin(15°) 2Sin(15°)=√3/2√2cot(x)+1/2√2 Sin(15°)=sin(45°-30°) =Sin(45°)cos(30°)-sin(30°)cos(45°) =(1/√2)(√3/2)-(1/√2)(1/2)=√3/2√2-1/2√2 2(√3/2√2-1/2√2)=√3/2√2cot(x)+1/2√2 2(√3-1)=√3cot(x)+1 √3cot(x)=2√3-2-1=2√3-3 cot(x)=(2√3-3)/√3 cot(x)=(6-3√3)/3=2-√3 So x=75°❤❤❤

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

Let's use an orthonormal, center D, first axis (DB) The equation of (BC) is y = -x +1, the equation of (DC) is y = -tan(60°).x = -sqrt(3).x. By intersection we have point C: C(-(1 + sqrt(3))/2; (3 +sqrt(3))/2) DC^2 = (1/4).(1 + 3 +2.sqrt(3) +9 + 3 +6.sqrt(3)) = 4 + 2.sqrt(3) VectorAC((3 -sqrt(3))/2; (3 +sqrt(3))/2) and AC^2 = (1/4).(9 + 3 -6.sqrt(3) + 9 + 3 + 6.sqrt(3)) =6 In trianglesADC: CD^2 = AC^2 + AD^2 - 2.AC.AD.cos(x) or 4 + 2.sqrt(3) = 6 + 4 - 4.sqrt(6).cos(x), then cos(x) = (6 - 2.sqrt(3))/(4.sqrt(6)) = (sqrt(6) - sqrt(2))/4 = cos(15°) = sin(75°) Finally x = 75°

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

As phungpham1725 did, drop a perpendicular from C to AB and label the intersection as point H. Let DH = a. ΔCDH is a 30°-60°-90° special right triangle, with longer side √3 times as long as the short side, so CH = a√3. ΔCBH is a 45°-45°-90° special isosceles right triangle, so its sides CH and BH are equal. BH = a + 1 so a + 1 = a√3, and a = 1/(√3 - 1). AH = 2 - a = 2 - 1/(√3 - 1) = (2√3 - 2)/(√3 - 1) - 1/(√3 - 1) = (2√3 - 3)/(√3 - 1). CH = a√3 = (1/(√3 - 1))(√3) = √3/(√3 - 1). tan(x) = CH/AH = (√3/(√3 - 1))/((2√3 - 3)/(√3 - 1)) = (√3)/((2√3 - 3)) = 3/(6 - 3√3) = 1/(2 - √3), and arctan(1/(2 - √3)) = 75° to within the precision of my scientific calculator. If we are not required to provide an exact result, we are done. Otherwise: The 15°-75°-90° right triangle appears frequently in geometry problems, so we check for ΔACH being such a triangle, in which case, x would be exactly 75°. We recall that the sides of a 15°-75°-90° right triangle, short - long - hypotenuse, are (√3 - 1):(√3 + 1):2√2, or (long)/(short) = (√3 + 1)/(√3 - 1) and check for a match. Since (2 - √3) is smaller than 1, we multiply both numerator and denominator by (√3 + 1) to make the numerator equal (√3 + 1) and see if the denominator is (√3 - 1). So denominator (2 - √3)(√3 + 1) = 2(√3 + 1) - (√3)(√3 + 1) = 2√3 + 2 - (√3)(√3) - (√3)(1) = 2√3 + 2 - (3) - √3 = (√3 - 1). We have the same numerator and denominator as for the 15°-75°-90° right triangle. Angle x is opposite the long side, so x = 75° exactly, as PreMath also found.

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

What an elegant solution. Mine involved the use of the sine rule and lots of fiddly algebra. Much prefer yours.

0:44

Nice geometric method.

@PreMath

11 күн бұрын

Glad you think so! Thanks for the feedback ❤️

Drop a perpendicular from C to AB in E EBC is isoceles right triangle so Let AE= x BE= 3-x We have AEC is a right triangle So AC^2= x^2 + (3-x)^2 = 9-6x^2 BC^2 = 2(3-x)^2 = 18-12x+2x^2 In ABC we have all the sides know so let's use Law of cosines cos x = AC^2+AB^2-BC^2/2*AC*BC 9+9-6x^2-18+12x-2x^2/2*√(18-12x+2x^2) √(9-6x^2) Simplifies to (√6-√2)/4 which is cos 75° So x=75°

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

Apply sine rule to CBD: CD = sin(45)/sin(15) = 2.73 Apply sine rule to ACD: sin(x)/CD = sin(120-x)/2 = (√3cos(x) + Sin(x))/4 => 4/CD = √3cot(x) +1 Giving cot(x) = 0.27 and x = 75°.

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

I used both the sine and cosine formulae to determine the size of x. It takes less time.

I was so close to the solution 😭 next time I will be able to solve these problems

@PreMath

12 күн бұрын

Great attitude! Keep it up👍 Thanks for the feedback ❤️

E is the circumcenter of ABC.😮

@PreMath

11 күн бұрын

Thanks for the feedback ❤️

What an amazing problem. I suppose the intuition between drawing the perpendicular would have been that the sides were of lengths 2 and 1. I was imaging how the problem would vary with different lengths but I should’ve focused on the fact that the lengths were 2 and 1. I guess that could be something to look out for in other problems, maybe it will help! Trigonometry solution : easy, geometry solution : 😮

x=75°

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

Let's find x: . .. ... .... ..... Let's assume that A is the center of the coordinate system and that AB is located on the x-axis. Then we obtain the following coordina tes: A: ( 0 ; 0 ) B: ( 3 ; 0 ) C: ( xC ; yC ) D: ( 2 ; 0 ) The lines CB and CD can be represented by the following functions: CB: y = −tan(45°)*(x − 3) = −(x − 3) CD: y = −tan(60°)*(x − 2) = −√3*(x − 2) Both lines intersect at C, so we can conclude: −(xC − 3) = −√3*(xC − 2) xC − 3 = √3*(xC − 2) xC − 3 = √3*xC − 2√3 2√3 − 3 = √3*xC − xC 2√3 − 3 = (√3 − 1)*xC xC = (2√3 − 3)/(√3 − 1) = (2√3 − 3)(√3 + 1)/[(√3 − 1)(√3 + 1)] = (6 + 2√3 − 3√3 − 3)/(3 − 1) = (3 − √3)/2 yC = −(xC − 3) = 3 − xC = 3 − (3 − √3)/2 = 6/2 − (3 − √3)/2 = (6 − 3 + √3)/2 = (3 + √3)/2 Now we are able to calculate x: tan(x) = (yC − yA)/(xC − xA) = yC/xC = [(3 + √3)/2]/[(3 − √3)/2] = (3 + √3)/(3 − √3) = (1 + 1/√3)/(1 − 1/√3) = (1 + 1/√3)/[1 − 1*(1/√3)] = [tan(45°) + tan(30°)]/[1 − tan(45°)*tan(30°)] = tan(45° + 30°) = tan(75°) ⇒ x = 75° Best regards from Germany

@unknownidentity2846

12 күн бұрын

After viewing the video I have to say that I like your solution very much.👍

@LuisdeBritoCamacho

12 күн бұрын

I like your Way of Reasoning. Congratulations!!

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

ctgx=(4sin15/√3sin45)-1/√3...tgx=2+√3...x=75

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

As ∠ADC is an exterior angle to ∆DBC at D, ∠BCD = ∠ADC-∠DBC = 60°-45° = 15°. Additionally, as ∠ADB is a straight angle, ∠CDB = 180°-60° = 120°. As ∠CDB is an exterior angle to ∆DCA at D, ∠DCA = ∠CDB-∠CAD = 120°-x. By the law of sines: 1/sin(15°) = BC/sin(120°) BC = sin(120°)/sin(15°) sin(15°) = sin(60°-45°) sin(15°) = sin(60°)cos(45°) - cos(60°)sin(45°) sin(15°) = (√3/2)(1/√2) - (1/2)(1/√2) sin(15°) = (√3-1)/2√2 BC = (√3/2)/((√3-1)/2√2) BC = (√3/2)(2√2)/(√3-1) = √6/(√3-1) BC = √6(√3+1)/(√3-1)(√3+1) BC = (3√2+√6)/(3-1) BC = (3√2+√6)/2 By the law of cosines: CA² = AB² + BC² - 2(AB)(BC)cos(45°) CA² = 3² + ((3√2+√6)/2)² - 2(3)((3√2+√6)/2)/√2 CA² = 9 + (√3+1)²(6/4) - 3(3√2+√6)/√2 CA² = 9 + (3/2)(4+2√3) - (9+3√3) CA² = 3(2+√3) - 3√3 = 6 CA = √6 By the law of sines: 2/sin(120°-x) = √6/sin(60°) √6sin(120°-x) = 2sin(60°) = 2(√3/2) sin(120°-x) = √3/√6 = 1/√2 = sin(45°) 120° - x = 45° x = 120° - 45° = 75°

@PreMath

11 күн бұрын

Excellent! Thanks for sharing ❤️

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Draw a Vertical Line from Point C until Red Line. Call this Point E. CE = h 02) Draw two Lines : One Horizontal passing through Point C and another Vertical passing through Point B. The Meeting Point is F 03) Angle DBC = 45º 04) Angle BDC = 120º 05) Angle DCB = 180º - (120º + 45º) = 180º - 165º = 15º 06) Angle ECD = 180º - (90º + 60º) = 180º - 150º = 30º 07) Now : Angle ECB = 45º 08) Line CB is the Diagonal of a Square [BECF]. 09) Triangle CDE is (30º - 60º - 90º) 10) Calculating the Diameter (D) of Square [BECF] : 11) 1 / sin(15º) = D / sin(120º) ; D = sin(120º) / sin(15º) ; D ~ 3,35 (Exact Form = sqrt(6) / (sqrt(3) - 1) 12) D = S * sqrt(2). Being "S" the Side Length of Square [BECF] 13) 3,35 = S * sqrt(2) ; S ~ 2,366 (Exact Form = sqrt(3) / (sqrt(3) - 1) 14) AE = 3 - 2.366 ~ 0,634 15) X = arctan(2,366 / 0,634) ; X ~ 74,999º 16) ANSWER : The Angle X is equal to 75º 17) A little bit boring! Best Wishes from Cordoba Caliphate - Center for the Studies of Divine Knowledge, Thinking and Wisdom. Department of Mathematics.

@PreMath

11 күн бұрын

Excellent! You are the best!🌹 Thanks for sharing ❤️

45 degree?

I want $100 dollars for every degree. $7,500 dollars. Thank you very much. 🙂

1$t view

Pin me 😂😂

@PreMath

12 күн бұрын

Thanks dear❤️

0:44