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Question-192001

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Question Number 192001 by ajfour last updated on 05/May/23
Commented by ajfour last updated on 05/May/23
Find x, in terms of a, b, m=tan θ.
Findx,intermsofa,b,m=tanθ.
Answered by ajfour last updated on 05/May/23
say θ=2φ, x=c cos (φ+sin^(−1) ((a−c)/(a+c))+sin^(−1) ((a−b)/(a+b))+φ) =(((a+b)^2 −(b+c)^2 +(c+a)^2 )/(2(a+b)(c+a))) say if φ=0 (((2(√(ac)))/(a+c)))(((2(√(ab)))/(a+b)))−(((a−c)/(a+c)))(((a−b)/(a+b))) =(((a+b)^2 −(b+c)^2 +(c+a)^2 )/(2(a+b)(c+a))) ⇒ 4a(√(bc))−{a^2 −(b+c)a+bc} =a^2 +a(b+c)−bc ⇒ 2a^2 −4a(√(bc))=0 ⇒ if a≠0 then a=2(√(bc)) ✓
sayθ=2ϕ,x=ccos(ϕ+sin−1a−ca+c+sin−1a−ba+b+ϕ)=(a+b)2−(b+c)2+(c+a)22(a+b)(c+a)sayifϕ=0(2aca+c)(2aba+b)−(a−ca+c)(a−ba+b)=(a+b)2−(b+c)2+(c+a)22(a+b)(c+a)⇒4abc−{a2−(b+c)a+bc}=a2+a(b+c)−bc⇒2a2−4abc=0⇒ifa≠0thena=2bc✓
Answered by mr W last updated on 05/May/23
Commented by mr W last updated on 05/May/23
cos α=((a−b)/(a+b)) cos β=((a−x)/(a+x)) cos γ=(((a+b)^2 +(a+x)^2 −(b+x)^2 )/(2(a+b)(a+x)))=((a(a+b)+(a−b)x)/(a(a+b)+(a+b)x)) α+β+γ=180+θ γ=180+θ−(α+β) let δ=α−θ=cos^(−1) ((a−b)/(a+b))−θ cos γ=−cos (δ+β)=−cos δ cos β+sin δ sin β ⇒((a(a+b)+(a−b)x)/(a(a+b)+(a+b)x))+(((a−x)/(a+x))) cos δ−sin δ (√(1−(((a−x)/(a+x)))^2 ))=0
cosα=a−ba+bcosβ=a−xa+xcosγ=(a+b)2+(a+x)2−(b+x)22(a+b)(a+x)=a(a+b)+(a−b)xa(a+b)+(a+b)xα+β+γ=180+θγ=180+θ−(α+β)letδ=α−θ=cos−1a−ba+b−θcosγ=−cos(δ+β)=−cosδcosβ+sinδsinβ⇒a(a+b)+(a−b)xa(a+b)+(a+b)x+(a−xa+x)cosδ−sinδ1−(a−xa+x)2=0

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