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Question Number 202614 by a.lgnaoui last updated on 30/Dec/23
find x
findx
Commented by a.lgnaoui last updated on 30/Dec/23
Commented by a.lgnaoui last updated on 31/Dec/23
(x+5)sin 𝛟=7sin 𝛌+xsin 𝛟 (1) 5sin 𝛟 =7sin 𝛌 (2) AB^2 =AC^2 +BC^2 −2AC.BC.cos 𝛟 7^2 = (x+5cos 𝛟)^2 +(5sin 𝛟)^2 49=x^2 +10xcos 𝛟+25 24=x^2 +10xcos 𝛟 ; cos 𝛟=((24−x^2 )/(10x)) 49=x^2 +(x+5)^2 −2x(x+5)cos 𝛟 x^3 +15x^2 +26x−240=0 ⇒ x=3
(x+5)sin\boldsymbolφ=7sin\boldsymbolλ+xsin\boldsymbolφ(1)5sin\boldsymbolφ=7sin\boldsymbolλ(2)\boldsymbolAB2=\boldsymbolAC2+\boldsymbolBC2−2\boldsymbolAC.\boldsymbolBC.cos\boldsymbolφ72=(\boldsymbolx+5cos\boldsymbolφ)2+(5sin\boldsymbolφ)249=x2+10\boldsymbolxcos\boldsymbolφ+2524=\boldsymbolx2+10\boldsymbolxcos\boldsymbolφ;cos\boldsymbolφ=24−\boldsymbolx210\boldsymbolx49=\boldsymbolx2+(\boldsymbolx+5)2−2\boldsymbolx(\boldsymbolx+5)cos\boldsymbolφ\boldsymbolx3+15\boldsymbolx2+26\boldsymbolx−240=0⇒\boldsymbolx=3
Commented by a.lgnaoui last updated on 31/Dec/23
Answered by mr W last updated on 31/Dec/23
cos α=−cos β ((x^2 +(x+5)^2 −7^2 )/(2x(x+5)))=−((x^2 +5^2 −7^2 )/(2×5x)) x^3 +15x^2 +26x−240=0 (x−3)(x+8)(x+10)=0 ⇒x=3 ✓
cosα=−cosβx2+(x+5)2−722x(x+5)=−x2+52−722×5xx3+15x2+26x−240=0(x−3)(x+8)(x+10)=0⇒x=3✓
Commented by mr W last updated on 31/Dec/23
Answered by AST last updated on 30/Dec/23
7^2 =5^2 +x^2 −10xcos(θ)⇒24=x^2 −10xcos(θ) 7^2 =x^2 +(5+x)^2 −2x(5+x)cos(180−θ) ⇒49=x^2 +25+10x+x^2 +10xcos(θ)+2x^2 cos(θ) ⇒24=2x^2 +10x+10xcos(θ)+2x^2 cos(θ) ⇒x+10+20cos(θ)+2xcos(θ)=0 x(1+2cos(θ))=−10(1+2cos(θ) x=−10 or cos(θ)=((−1)/2) ⇒x^2 +5x−24=0⇒x=3 [θ is the angle between length (5+x) and (x)]
72=52+x2−10xcos(θ)⇒24=x2−10xcos(θ)72=x2+(5+x)2−2x(5+x)cos(180−θ)⇒49=x2+25+10x+x2+10xcos(θ)+2x2cos(θ)⇒24=2x2+10x+10xcos(θ)+2x2cos(θ)⇒x+10+20cos(θ)+2xcos(θ)=0x(1+2cos(θ))=−10(1+2cos(θ)x=−10orcos(θ)=−12⇒x2+5x−24=0⇒x=3[θistheanglebetweenlength(5+x)and(x)]

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