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Show-that-in-a-30-60-90-triangle-the-altitude-on-the-hypotaneuse-divides-the-hypotaneuse-into-segments-whose-length-has-the-ratio-1-3-without-using-trigonometry-




Question Number 60043 by Kunal12588 last updated on 17/May/19
Show that in a 30°−60°−90° triangle the   altitude on the hypotaneuse divides the   hypotaneuse into segments whose length  has the ratio 1/3.  without using trigonometry.
Showthatina30°60°90°trianglethealtitudeonthehypotaneusedividesthehypotaneuseintosegmentswhoselengthhastheratio1/3.withoutusingtrigonometry.
Answered by tanmay last updated on 17/May/19
The eqn of hypotaneous  (x/a)+(y/b)=1  bx+ay=ab  y=((−bx)/a)+b →slopse m_1 =−(b/a)  eq of st line ⊥to hypotaneous and passing  through origin(0,0)  y=m_2 x    m_1 ×m_2 =−1  m_2 =((−1)/((−b)/a))=(a/b)     y=(a/b)x  solve y=(a/b)x and (x/a)+(y/b)=1  to get point C which devide AB   in (1/3)  ratio  (x/b)=(y/a)=k→x=bk   y=ak    ((bk)/a)+((ak)/b)=1→k(a^2 +b^2 )=ab  k=((ab)/(a^2 +b^2 ))  x=((ab^2 )/(a^2 +b^2 ))   y=((a^2 b)/(a^2 +b^2 ))  A(0,b)   B(a,0)   O(0,0)   and C(((ab^2 )/(a^2 +b^2 )),((a^2 b)/(a^2 +b^2 )))  let C devides AB    (m/n) ratio  ((ab^2 )/(a^2 +b^2 ))=((m×0+n×a)/(m+n))  mab^2 +nab^2 =na^3 +nab^2   (m/n)=(a^3 /(ab^2 ))=(a^2 /b^2 )  now slope of AB is m_1 =((−b)/a)  let AB makes angle θ with OB  so  slope(m_1 )=((−b)/a)=tan(180^o −60^o )=−(√3)   (b/a)=(√3)   (m/n)=(a^2 /b^2 )=(1/3)  proved  ■■ if we take θ=30^o   slope=((−b)/a)=tan(180^o −30^o )=−(1/( (√3)))  (b/a)=(1/( (√3)))  then (m/n)=(a^2 /b^2 )=3   so rstion (3/1) also fulfill...    note Triangle OAB  OA⊥OB   znd AB hypotaneous
Theeqnofhypotaneousxa+yb=1bx+ay=aby=bxa+bslopsem1=baeqofstlinetohypotaneousandpassingthroughorigin(0,0)y=m2xm1×m2=1m2=1ba=aby=abxsolvey=abxandxa+yb=1togetpointCwhichdevideABin13ratioxb=ya=kx=bky=akbka+akb=1k(a2+b2)=abk=aba2+b2x=ab2a2+b2y=a2ba2+b2A(0,b)B(a,0)O(0,0)andC(ab2a2+b2,a2ba2+b2)letCdevidesABmnratioab2a2+b2=m×0+n×am+nmab2+nab2=na3+nab2mn=a3ab2=a2b2nowslopeofABism1=baletABmakesangleθwithOBsoslope(m1)=ba=tan(180o60o)=3ba=3mn=a2b2=13proved◼◼ifwetakeθ=30oslope=ba=tan(180o30o)=13ba=13thenmn=a2b2=3sorstion31alsofulfillnoteTriangleOABOAOBzndABhypotaneous
Commented by Kunal12588 last updated on 17/May/19
sir you still used trigonometry (tangent)
siryoustillusedtrigonometry(tangent)
Commented by Kunal12588 last updated on 17/May/19
but i like your way sir (pls let it be here)
butilikeyourwaysir(plsletitbehere)
Answered by mr W last updated on 17/May/19
Commented by mr W last updated on 17/May/19
let AB=1, AF=x  ⇒BC=CE=1  ⇒CD=x  ⇒FE=ED=(√(x^2 −1))  ⇒AD=2+x  AD^2 =AF^2 +FD^2   ⇒(2+x)^2 =x^2 +(2(√(x^2 −1)))^2   ⇒4+4x+x^2 =x^2 +4x^2 −4  ⇒x^2 −x−2=0  ⇒(x+1)(x−2)=0  ⇒x=2  ⇒BD=1+2=3  ⇒AB/BD=1/3 ⇒proved
letAB=1,AF=xBC=CE=1CD=xFE=ED=x21AD=2+xAD2=AF2+FD2(2+x)2=x2+(2x21)24+4x+x2=x2+4x24x2x2=0(x+1)(x2)=0x=2BD=1+2=3AB/BD=1/3proved
Commented by Kunal12588 last updated on 17/May/19
sir i tried pls check  △AFC is equilateral   ⇒FC=AC=AB+BC=2  FC=DC  ⇒AC=DC  ⇒x=2  ((AB)/(BD))=(1/(x+1))=(1/(2+1))=(1/3)
siritriedplscheckAFCisequilateralFC=AC=AB+BC=2FC=DCAC=DCx=2ABBD=1x+1=12+1=13
Commented by mr W last updated on 17/May/19
nice path sir!
nicepathsir!

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