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Question Number 47282 by behi83417@gmail.com last updated on 07/Nov/18

Answered by MrW3 last updated on 07/Nov/18

Commented by MrW3 last updated on 08/Nov/18

AB=c  AD=DB=(c/2)  CD=m_c   let λ=(c/(2m_c ))  (m_c /(sin A))=(c/(2 sin α))  ⇒sin α=(c/(2m_c )) sin A=λ sin A  similarly  sin β=λ sin B  α+β=∠C=π−(∠A+∠B)  ⇒sin (α+β)=sin (A+B)  ⇒sin α cos β+cos α sin β=sin (A+B)  ⇒λ(sin A (√(1−λ^2  sin^2  B))+sin B (√(1−λ^2  sin^2  A)))=sin (A+B)  A=20°, B=45°  A+B=65°≠90°  ⇒λ=1 (not suitable since A+B≠90°)  ⇒λ=1.4112  ⇒c=2×1.4112×(√5)=6.3111  (h/(tan A))+(h/(tan B))=c  ⇒h=(c/(cot A+cot B))=1.6841  A_(ΔABC) =(1/2)ch=(1/2)×6.3111×1.6841  =5.314.

AB=cAD=DB=c2CD=mcletλ=c2mcmcsinA=c2sinαsinα=c2mcsinA=λsinAsimilarlysinβ=λsinBα+β=C=π(A+B)sin(α+β)=sin(A+B)sinαcosβ+cosαsinβ=sin(A+B)λ(sinA1λ2sin2B+sinB1λ2sin2A)=sin(A+B)A=20°,B=45°A+B=65°90°λ=1(notsuitablesinceA+B90°)λ=1.4112c=2×1.4112×5=6.3111htanA+htanB=ch=ccotA+cotB=1.6841AΔABC=12ch=12×6.3111×1.6841=5.314.

Commented by behi83417@gmail.com last updated on 07/Nov/18

thanks in advance dear master.

thanksinadvancedearmaster.

Answered by behi83417@gmail.com last updated on 07/Nov/18

Commented by MrW3 last updated on 08/Nov/18

nice method sir!

nicemethodsir!

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