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

Very nice problem.

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 😅

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

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°❤❤❤

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°

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

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.

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°

Interesting

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°.

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

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°

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

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.

Elegant Solution

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

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

Nice geometric method.

Thank you!

x=75°

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

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

E is the circumcenter of ABC.😮

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 : 😮

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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.

45 degree?

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

1$t view

Pin me 😂😂

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