Can you find the value of m/n? | (Justify) | 

Very cool question!! 👍👏

Without trigonometry. Additional constructions: 1. ∠ВАК = х (К - on ВС). 2. DM - мedian in ▲DCВ. 3. КН ⟂ АВ, МР ⟂ АВ. СК = 16, МВ = 18, КМ = 2. КМ : МВ = НР : РВ = НР : DР = 1 : 9. НD = (8/9)DР = (8/9)n/2. АН = (m + n)/2. АD = m = АН - НD = (m + n)/2 - (8/9)n/2. 9m + 9n - 8n = 18m. n = 9m, m/n = 1/9.

16sin3x=36sinx...(sinx)^2=3/16..m=16cos3x=16cosx(1-4(sinx)^2)=16cosx*(1/4)..n=36cosx..m/n=(16/36)(1/4)=1/9

wow!

Let's face this challenge: . .. ... .... ..... The triangles ACD and BCD are both right triangles. Therefore we can conclude: sin(3x) = CD/AC ⇒ CD = AC*sin(3x) sin(x) = CD/BC ⇒ CD = BC*sin(x) AC*sin(3x) = BC*sin(x) sin(3x)/sin(x) = BC/AC = 36/16 = 9/4 sin(3x) = sin(2x + x) = sin(2x)*cos(x) + cos(2x)*sin(x) = 2*sin(x)*cos(x)*cos(x) + [cos²(x) − sin²(x)]*sin(x) = 2*sin(x)*cos²(x) + sin(x)*cos²(x) − sin³(x) = 3*sin(x)*cos²(x) − sin³(x) = 3*sin(x)*[1 − sin²(x)] − sin³(x) = 3*sin(x) − 3*sin³(x) − sin³(x) = 3*sin(x) − 4*sin³(x) sin(3x)/sin(x) = 9/4 3 − 4*sin²(x) = 9/4 −4*sin²(x) = 9/4 − 3 = 9/4 − 12/4 = −3/4 sin²(x) = 3/16 ⇒ sin(x) = √3/4 CD = BC*sin(x) = 36*(√3/4) = 9√3 Now we can apply the Pythagorean theorem to the right triangles ACD and BCD: m² = AD² = AC² − CD² = 16² − (9√3)² = 256 − 243 = 13 n² = BD² = BC² − CD² = 36² − (9√3)² = 1296 − 243 = 1053 m²/n² = 13/1053 = 13/(13*81) = 1/81 ⇒ m/n = 1/9 Best regards from Germany

1/From A draw a line forming with AB an angle= x, intersecting BC at point E. We have 2 isosceles triangles: ACE and AEB. So AC=CE= 16 and AE=EB=20. 2/ Calculating cosine of angle ACE by cosine law: Sq20=sq16+sq16 -2 sq16. Cos(ACE) -> cos ACE=7/32 2/ Calculating AB by cosine law: SqAB= sq16+sq36 -2x16x36x7/32 -> AB= 10sqrt13 3/Calculating m and n: Label CD=h We have sq h= sq16-sq m and sq h= sq36 - sq(10sqrt13 - m) -> sq16-sqm=sq36-1300-sqm +20sqrt13.m -> m=sqrt13 and n=9sqrt13 The ratio m/n = 1/9 😅

My way of solution ▶ we write the Sinus theorem for the right triangle ΔADC: sin(3x)/h= sin(90-3x)/m= sin(90°)/16 sin(90°-3x)= sin(90°)*cos(3x) - cos(90°)*sin(3x) sin(90°- 3x)= cos(3x) ⇒ sin(90-3x)/m= sin(90°)/16 cos(x)/m= 1/16 ⇒ m= 16*cos(3x) b) we write the Sinus theorem for the right triangle ΔDBC: sin(x)/h= sin(90-x)/n= sin(90°)/36 sin(90°-x)= sin(90°)*cos(x) - cos(90°)*sin(x) sin(90°- x)= cos(x) ⇒ sin(90-x)/n= sin(90°)/36 cos(x)/n= 1/36 ⇒ n= 36*cos(x) c) m/n = 16cos(3x)/36cos(x) m/n = (16/36)*cos(3x)/cos(x) m/n = ........cos(2x+x)/cos(x) m/n = .........[cos(2x)*cos(x)-sin(2x)*sin(x)]/cos(x) sin(2x)= 2*sin(x)*cos(x) ⇒ m/n = .........[cos(2x) - 2sin(x)*cos(x)*sin(x)/cos(x)] m/n =..........[cos(2x) - 2sin²(x)] cos(2x)= 2cos²(x) -1 ⇒ m/n =..........[2cos²(x) -1 - 2sin²(x)] m/n = (16/36)[2cos²(x) -1 - 2sin²(x)].......................................Gl.-1 d) we write the Sinus theorem for the right triangle ΔABC: sin(3x)/36= sin(x)/16= sin(180°- 4x)/(m+n) ⇒ sin(3x)/36= sin(x)/16 sin(3x)/sin(x)= 36/16 sin(2x+x)/sin(x)= 9/4 [sin(2x)*cos(x)+cos(2x)*sin(x)]/sin(x)= 9/4 [2*sin(x)*cos(x)*cos(x) + cos(2x)*sin(x)]/sin(x)= 9/4 [2cos²(x) + cos(2x)]= 9/4 cos(2x)= 2cos²(x)-1 ⇒ [2cos²(x) + 2cos²(x)-1]= 9/4 4cos²(x) -1= 9/4 4cos²(x)= 1 + 9/4 4cos²(x)= 13/4 cos²(x)= 13/16 cos(x)= √13/4 ⇒ sin²(x)= 1- 13/16 sin²(x)= 3/16 e) Equation-1 of step c : m/n= (16/36)[2cos²(x) -1 - 2sin²(x)] m/n= (4/9)[2*(13/16)-1 - 2*(3/16)] m/n= (4/9)[13/8 - 1 - 3/8] m/n= (4/9)[5/8 - 3/8] m/n= (4/9)(2/8) m/n= (4/9)*(1/4) m/n= 1/9 is the answer !

Very challenging, thanks PreMath. I did a cumbersome solution without trig. Construct CE with point E "m" distance to the right of D on AB. Bisect ECB with the bisector meeting AB at F. Triangle EBC is similar to EDF. And DF is equal to FB. Similar triangles gives DF and FB=72/sqrt13. Angle bisector theorem gives EF=32/sqrt13. Pythagoras on CDE and CDB gives m=sqrt13. Add DE, EF, and FB to get n=9*sqrt13.

Superb👍

Without trigonometric formulas I draw a length from A to a point (let’s say P) on BC sharing 3x into x and 2x then I saw that PC=AC since CPA=2x so CP=16 and since APB is an isocele if we draw a length from P to a point let’s say Q on DB so Q is the midpoint of DB. Now we have DQ=n-m/2 and QB=m+n/2 the result from now on is easy look at the triangle CDB use Thales theorem which gives us m/n=1/9!

According to euklid (in german, we call it "Kathetensatz"), m*(m+n)=16^2 and n*(m+n)=36^2. therefor 16^2/36^2=(m*(m+n))/(n*(m+n))=m/n. 16^2/36^2=(16/36)^2=(4/9)^2=16/81 Tthis result (m/n=16/81) is different from your result. Why? i just checked, if it is possibe, that the legs of the triangle are 16 and 36 and the anges are x and 3x, if m/n)=1/9. The hpothhenuse of the triangle is sqrt(16^2+36^2) which is about 39.4. If m/n=1/9, m must be about 3.94 and n must be about 35.46. inverse cosine of 3.94/16 is about 75.75°, inverse cosine of 35.46/36 is about 9.93°, so the angles of the triangle can not be x and 3x, if the legs of the rectangularttriangle are 16 and 36. I saw, it was my mistake: the triangle ABC is not a rectangular triangle ...

Right! ... so @ 4:12 & 8:00 those famous Triple Angle Identities look scary, but they're not really! They're just extensions of Double Angle Identities. The scary part about these Triple Nipple Aliens is that like Trigonometry having 6 Trigonometric Functions there are 6 Triple Angle Functions and committing them too memory takes 6 brains. 🙂

Let CD = y. In triangle ∆CDB, y = 36sin(x). In triangle ∆ADC, y = 16sin(3x). sin(3x) = sin(2x+x) sin(3x) = sin(2x)cos(x) + cos(2x)sin(x) sin(3x) = (2sin(x)cos(x))cos(x) + (cos²(x)-sin²(x))sin(x) sin(3x) = 2sin(x)cos²(x) + (1-sin²(x)-sin²(x))sin(x) sin(3x) = 2sin(x)(1-sin²(x)) + (1-2sin²(x))sin(x) sin(3x) = 2sin(x) - 2sin³(x) + sin(x) - 2sin³(x) sin(3x) = 3sin(x) - 4sin³(x) 36sin(x) = 16sin(3x) 36sin(x) = 16(3sin(x)-4sin³(x)) 36 = 16(3-4sin²(x)) 36 = 48 - 64sin²(x) 64sin²(x) = 12 sin²(x) = 12/64 = 3/16 sin(x) = √(3/16) = √3/4 y/36 = √3/4 36√3 = 4y y = 36√3/4 = 9√3 Triangle ∆CDB: CD² + DB² = BC² (9√3)² + n² = 36² 243 + n² = 1296 n² = 1296 - 243 = 1053 n = √1053 = 9√13 Triangle ∆ADC: AD² + CD² = CA² m² + 243 = 16² m² = 256 - 243 = 13 m = √13 m/n = √13/9√13 = 1/9

I would never find this solution. Does another one exist?

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Let CD = h 02) sin(3x) = h / 16 ; h = 16sin(3x) 03) sin(x) = h / 36 ; h = 36sin(x) 04) 16sin(3x) = 36sin(x) ; 4sin(x) = 9sin(3x) 05) x = 0,44783 radians ; x ~ 25,66º 06) cos (0,44783 rad) ~ 0,902 07) cos (3*0,4473 rad) ~ 0,225 08) n = 36*0,902 ; n ~ 32,450 09) m = 16*0,225 ; m ~ 3,606 10) m / n ~ 0,1111 Therefore, OUR ANSWER : m / n ~ 0,1111

Be h = CD. And tan(x) = t. *tan(3.x) = h/m and tan(x) = h/n, so m/n = tan(x)/tan(3.x) We know that tan(3.x) = (3.tan(x) - (tan(x))^3)/(1 - 3.(tan(x))^2) So m/n = (1 - 3.t^2)/(3 - t^2). *sin(3.x) = h/16 and sin(x) = h/36, so sin(3.x)/sin(x) = 9/4 We know that sin(3.x) = 3.sin(x) - 4.(sin(x))^3,, so 3 - 4.(sin(x)^2 = 9/4, and (sin(x))^2 =3/16. Then (cos(x))^2 = 1 - 3/16 = 13/16, and t^2 = (tan(x))^2 = 3/13 *Finally m/n = (1 -3.(3/13))/(3 - 3/13) = 4/36 = 1/9.

Geometric solution with Shorts: ru-vid.comlobbhqob-d4?si=50yR_Y8Ao3VwhX3o

m/n=1/9

The blurb says m/m, so I say the solution is 1 :) However, the video has m/n :)

asnwer=12cm

16/36=4/9(may be )

36 sin x=16 sin 3x, m=16 cos 3x, n=36 cos x, m/n=4/9 cos 3x/cos x,.....😅

sin3x = h/16 sinx = h/36 sin3x = sin(2x + x) = sin2x*cosx + sinx*cos2x h/16 = 2*sinx*cosx*cosx + sinx*(1 - 2sinx^2) h/16 = 2*sinx*cosx^2 + sinx*(1 - 2sinx^2) h/16 = 2*sinx*(1 - sinx^2) + sinx*(1 - 2sinx^2) h/16 = 2*h/36*(1 - (h/36)^2) + h/36*(1 - 2*(h/36)^2) h/16 = h/18*(1 - h^2/1296) + h/36*(1 - 2*(h^2/1296)) h/16 = h/18*(1 - h^2/1296) + h/36*(1 - h^2/648) h/16 = h/18*((1296 - h^2)/1296) + h/36*((648 - h^2)/648) 1458*h = h*(1296 - h^2) + h*(648 - h^2) 1458*h = 1296*h - h^3 + 648*h - h^3 2h^3 = 486*h h^2 = 243 16^2 = h^2 + m^2 m^2 = 256 - 243 m^2 = 13 m = sqrt(13) 36^2 = h^2 + n^2 n^2 = 1296 - 243 n^2 = 1053 n = sqrt(1053) n = 9*sqrt(13) m/n = sqrt(13)/9*sqrt(13) m/n = 1/9