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Question-117480




Question Number 117480 by I want to learn more last updated on 12/Oct/20
Commented by Tawa11 last updated on 15/Sep/21
nice
nice
Answered by mr W last updated on 12/Oct/20
Commented by mr W last updated on 12/Oct/20
let BC=a  cos β=((21)/a)  cos γ=((35)/a)  ((AC)/(sin β))=((BC)/(sin (β+γ)))=(a/(sin β cos γ+cos β sin γ))  AC=((a sin β)/(sin β cos γ+cos β sin γ))  DC=AC cos γ=((a sin β cos γ)/(sin β cos γ+cos β sin γ))  DG tan β=DC  15 tan β =((a sin β cos γ)/(sin β cos γ+cos β sin γ))  15=((a cos β cos γ)/(sin β cos γ+cos β sin γ))  tan β+tan γ=(a/(15))  (√(((a/(21)))^2 −1))+(√(((a/(35)))^2 −1))=(a/(15))  ((a/(21)))^2 −1=((a/(15)))^2 +((a/(35)))^2 −1−2((a/(15)))(√(((a/(35)))^2 −1))  a^2 [(1/(35^2 ))−((1/(15^2 ))+(1/(35^2 ))−(1/(21^2 )))^2 (((15)/2))^2 ]=1  ⇒a=(1/( (√((1/(35^2 ))−((1/(15^2 ))+(1/(35^2 ))−(1/(21^2 )))^2 (((15)/2))^2 ))))=((98)/( (√3)))  ⇒β=cos^(−1) ((21)/a)=cos^(−1) ((3(√3))/(14))  ⇒γ=cos^(−1) ((35)/a)=cos^(−1) ((5(√3))/(14))  ⇒α=π−cos^(−1) ((3(√3))/(14))−cos^(−1) ((5(√3))/(14))=(π/3)=60°
letBC=acosβ=21acosγ=35aACsinβ=BCsin(β+γ)=asinβcosγ+cosβsinγAC=asinβsinβcosγ+cosβsinγDC=ACcosγ=asinβcosγsinβcosγ+cosβsinγDGtanβ=DC15tanβ=asinβcosγsinβcosγ+cosβsinγ15=acosβcosγsinβcosγ+cosβsinγtanβ+tanγ=a15(a21)21+(a35)21=a15(a21)21=(a15)2+(a35)212(a15)(a35)21a2[1352(1152+13521212)2(152)2]=1a=11352(1152+13521212)2(152)2=983β=cos121a=cos13314γ=cos135a=cos15314α=πcos13314cos15314=π3=60°
Commented by I want to learn more last updated on 12/Oct/20
I appreciate sir.
Iappreciatesir.

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