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Question Number 149894 by ajfour last updated on 08/Aug/21

Commented by ajfour last updated on 08/Aug/21

Find minimum and maximum  values for the side of equilateral  triangle shown.

Findminimumandmaximumvaluesforthesideofequilateraltriangleshown.

Answered by mr W last updated on 08/Aug/21

Commented by mr W last updated on 08/Aug/21

Commented by mr W last updated on 08/Aug/21

P(h+r cos θ,k+r sin θ)  eqn. of AP:  y=k+r sin θ+m_1 x  eqn. of BP:  y=k+r sin θ+mx  tan (π/3)=((m_1 −m)/(1+m_1 m))=(√3)  ⇒m_1 =((m+(√3))/(1−(√3)m))  s=(h+r cos θ)(√(1+m^2 ))  s=(k+r sin θ)(√(1+(1/m_1 ^2 )))  with α=(√(1+m^2 )), β=(√(1+(1/m_1 ^2 )))  (h+r cos θ)α=(k+r sin θ)β  r(β sin θ−α cos θ)=αh−βk  r(√(α^2 +β^2 )) sin (θ−tan^(−1) (α/β))=αh−βk  sin (θ−tan^(−1) (α/β))=((αh−βk)/(r(√(α^2 +β^2 ))))  ⇒θ=tan^(−1) (α/β)+sin^(−1) (((αh−βk)/(r(√(α^2 +β^2 ))))) or  ⇒θ=tan^(−1) (α/β)+π−sin^(−1) (((αh−βk)/(r(√(α^2 +β^2 )))))

P(h+rcosθ,k+rsinθ)eqn.ofAP:y=k+rsinθ+m1xeqn.ofBP:y=k+rsinθ+mxtanπ3=m1m1+m1m=3m1=m+313ms=(h+rcosθ)1+m2s=(k+rsinθ)1+1m12withα=1+m2,β=1+1m12(h+rcosθ)α=(k+rsinθ)βr(βsinθαcosθ)=αhβkrα2+β2sin(θtan1αβ)=αhβksin(θtan1αβ)=αhβkrα2+β2θ=tan1αβ+sin1(αhβkrα2+β2)orθ=tan1αβ+πsin1(αhβkrα2+β2)

Commented by ajfour last updated on 08/Aug/21

sir m, m_1   stand undetermined.

sirm,m1standundetermined.

Commented by mr W last updated on 08/Aug/21

i tried to find the length s in terms of  m in following way:  m_1 =((m+(√3))/(1−(√3)m))  α=(√(1+m^2 )), β=(√(1+(1/m_1 ^2 )))  θ=tan^(−1) (α/β)+sin^(−1) (((αh−βk)/(r(√(α^2 +β^2 ))))) or  θ=tan^(−1) (α/β)+π−sin^(−1) (((αh−βk)/(r(√(α^2 +β^2 )))))  s=(h+r cos θ)(√(1+m^2 ))  then graphically find the maximum  and minimum of s.

itriedtofindthelengthsintermsofminfollowingway:m1=m+313mα=1+m2,β=1+1m12θ=tan1αβ+sin1(αhβkrα2+β2)orθ=tan1αβ+πsin1(αhβkrα2+β2)s=(h+rcosθ)1+m2thengraphicallyfindthemaximumandminimumofs.

Commented by mr W last updated on 08/Aug/21

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