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Question Number 137327 by mr W last updated on 01/Apr/21

in triangle ΔABC: BC=1, ∠B=2∠A. find the maximum area of ΔABC.

$${in}\:{triangle}\:\Delta{ABC}:\:{BC}=\mathrm{1},\:\angle{B}=\mathrm{2}\angle{A}. \\ $$$${find}\:{the}\:{maximum}\:{area}\:{of}\:\Delta{ABC}. \\ $$

Answered by EDWIN88 last updated on 01/Apr/21

∠A+∠B+∠C =π ; 3∠A+∠C=π let ∠A =α ; ∠B=2α ; ∠C=π−3α ((BC)/(sin α)) = ((AC)/(sin 2α)) ⇒ (1/(sin α)) = ((AC)/(2sin α cos α)) AC= 2cos α Area of ΔABC = (1/2).1.AC.sin (π−3α) A(α)=(1/2)(2 cos α sin 3α) A(α)=(1/2)(sin 4α+sin 2α) A′(α)=(1/2)(4cos 4α+2cos 2α)=0 ⇒ 2cos 4α+cos 2α = 0 ⇒2(2cos^2 2α−1)+cos 2α = 0 ⇒4cos^2 2α+cos 2α−2=0 cos 2α = ((−1+ (√(33)))/8) = 0.593 sin 2α =(√(1−(0.593)^2 )) = 0.805 Thus A(α)_(max) = (1/2)(2sin 2α cos 2α+sin 2α) A(α)_(max) = (1/2)sin 2α(2cos 2α+1) A(α)_(max) = (1/2)(0.805)(2.186)=0.8798

$$\angle\mathrm{A}+\angle\mathrm{B}+\angle\mathrm{C}\:=\pi\:;\:\mathrm{3}\angle\mathrm{A}+\angle\mathrm{C}=\pi \\ $$$$\mathrm{let}\:\angle\mathrm{A}\:=\alpha\:;\:\angle\mathrm{B}=\mathrm{2}\alpha\:;\:\angle\mathrm{C}=\pi−\mathrm{3}\alpha \\ $$$$\frac{\mathrm{BC}}{\mathrm{sin}\:\alpha}\:=\:\frac{\mathrm{AC}}{\mathrm{sin}\:\mathrm{2}\alpha}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{sin}\:\alpha}\:=\:\frac{\mathrm{AC}}{\mathrm{2sin}\:\alpha\:\mathrm{cos}\:\alpha} \\ $$$$\mathrm{AC}=\:\mathrm{2cos}\:\alpha\: \\ $$$$\mathrm{Area}\:\mathrm{of}\:\Delta\mathrm{ABC}\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{1}.\mathrm{AC}.\mathrm{sin}\:\left(\pi−\mathrm{3}\alpha\right) \\ $$$$\mathrm{A}\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\mathrm{3}\alpha\right) \\ $$$$\mathrm{A}\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{4}\alpha+\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$\mathrm{A}'\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4cos}\:\mathrm{4}\alpha+\mathrm{2cos}\:\mathrm{2}\alpha\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2cos}\:\mathrm{4}\alpha+\mathrm{cos}\:\mathrm{2}\alpha\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{2}\alpha−\mathrm{1}\right)+\mathrm{cos}\:\mathrm{2}\alpha\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{4cos}\:^{\mathrm{2}} \mathrm{2}\alpha+\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha\:=\:\frac{−\mathrm{1}+\:\sqrt{\mathrm{33}}}{\mathrm{8}}\:=\:\mathrm{0}.\mathrm{593} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha\:=\sqrt{\mathrm{1}−\left(\mathrm{0}.\mathrm{593}\right)^{\mathrm{2}} }\:=\:\mathrm{0}.\mathrm{805} \\ $$$$\mathrm{Thus}\:\mathrm{A}\left(\alpha\right)_{\mathrm{max}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2sin}\:\mathrm{2}\alpha\:\mathrm{cos}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$\mathrm{A}\left(\alpha\right)_{\mathrm{max}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\alpha\left(\mathrm{2cos}\:\mathrm{2}\alpha+\mathrm{1}\right) \\ $$$$\mathrm{A}\left(\alpha\right)_{\mathrm{max}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}.\mathrm{805}\right)\left(\mathrm{2}.\mathrm{186}\right)=\mathrm{0}.\mathrm{8798} \\ $$

Commented by mr W last updated on 01/Apr/21

yes. the exact value is ((((√(33))+3)(√(((√(33))+3)((√(33))−1))))/(64))≈0.8801

$${yes}.\:{the}\:{exact}\:{value}\:{is} \\ $$$$\frac{\left(\sqrt{\mathrm{33}}+\mathrm{3}\right)\sqrt{\left(\sqrt{\mathrm{33}}+\mathrm{3}\right)\left(\sqrt{\mathrm{33}}−\mathrm{1}\right)}}{\mathrm{64}}\approx\mathrm{0}.\mathrm{8801} \\ $$

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