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Question Number 68425 by mr W last updated on 10/Sep/19

Commented by Prithwish sen last updated on 10/Sep/19

AB=2x AC=x 3x×100×((√3)/2) =2x^2 ×((√3)/2) 2x^2 −300x=0 ⇒x=150 AB=300 AC=150 BC =(√(300^2 +150^2 −2×300×150×cos120)) = (√(90000+22500+45000)) = (√(157500))=150(√7)

$$\mathrm{AB}=\mathrm{2x}\:\:\mathrm{AC}=\mathrm{x} \\ $$$$\mathrm{3x}×\mathrm{100}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{2x}^{\mathrm{2}} ×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{2x}^{\mathrm{2}} −\mathrm{300x}=\mathrm{0}\:\Rightarrow\mathrm{x}=\mathrm{150} \\ $$$$\mathrm{AB}=\mathrm{300}\:\:\mathrm{AC}=\mathrm{150} \\ $$$$\mathrm{BC}\:=\sqrt{\mathrm{300}^{\mathrm{2}} +\mathrm{150}^{\mathrm{2}} −\mathrm{2}×\mathrm{300}×\mathrm{150}×\mathrm{cos120}} \\ $$$$\:\:\:\:\:\:\:\:=\:\sqrt{\mathrm{90000}+\mathrm{22500}+\mathrm{45000}} \\ $$$$\:\:\:\:\:\:\:\:=\:\sqrt{\mathrm{157500}}=\mathrm{150}\sqrt{\mathrm{7}} \\ $$

Commented by mr W last updated on 10/Sep/19

thanks alot sir!

$${thanks}\:{alot}\:{sir}! \\ $$

Answered by ajfour last updated on 10/Sep/19

let AB=2AC=2p BD=x, DC=y cos 120°=((p^2 +4p^2 −(x+y)^2 )/(4p^2 ))=−(1/2) ⇒ (x+y)^2 =7p^2 ((BD)/(DC))=(x/y)=((2p)/p)=2 ⇒ 3y=p(√7) or y=((p(√7))/3) , x=((2p(√7))/3) considering △ADC cos 60°=((AD^2 +AC^2 −CD^2 )/(2(AD)(AC)))=(1/2) say AD=a ((a^2 +p^2 −y^2 )/(2ap))=(1/2) a^2 +p^2 −((7p^2 )/9)=ap ((2p^2 )/9)−ap+a^2 =0 ⇒ 2p^2 −9ap+9a^2 =0 2p^2 −6ap−3ap+9a^2 =0 2p(p−3a)−3a(p−3a)=0 p=((3a)/2), or p=3a (i should reject) BC=x+y = p(√7) = (((3(√7))/2))a BC = 150(√7) .

$${let}\:{AB}=\mathrm{2}{AC}=\mathrm{2}{p}\:\: \\ $$$$\:{BD}={x},\:{DC}={y} \\ $$$$\mathrm{cos}\:\mathrm{120}°=\frac{{p}^{\mathrm{2}} +\mathrm{4}{p}^{\mathrm{2}} −\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{4}{p}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{7}{p}^{\mathrm{2}} \\ $$$$\:\:\frac{{BD}}{{DC}}=\frac{{x}}{{y}}=\frac{\mathrm{2}{p}}{{p}}=\mathrm{2} \\ $$$$\Rightarrow\:\:\mathrm{3}{y}={p}\sqrt{\mathrm{7}}\:\:{or}\:\:{y}=\frac{{p}\sqrt{\mathrm{7}}}{\mathrm{3}}\:,\:{x}=\frac{\mathrm{2}{p}\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$$\:\:{considering}\:\bigtriangleup{ADC} \\ $$$$\:\:\:\:\mathrm{cos}\:\mathrm{60}°=\frac{{AD}^{\mathrm{2}} +{AC}^{\mathrm{2}} −{CD}^{\mathrm{2}} }{\mathrm{2}\left({AD}\right)\left({AC}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${say}\:{AD}={a} \\ $$$$\:\:\:\frac{{a}^{\mathrm{2}} +{p}^{\mathrm{2}} −{y}^{\mathrm{2}} }{\mathrm{2}{ap}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:{a}^{\mathrm{2}} +{p}^{\mathrm{2}} −\frac{\mathrm{7}{p}^{\mathrm{2}} }{\mathrm{9}}={ap} \\ $$$$\frac{\mathrm{2}{p}^{\mathrm{2}} }{\mathrm{9}}−{ap}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{2}{p}^{\mathrm{2}} −\mathrm{9}{ap}+\mathrm{9}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\mathrm{2}{p}^{\mathrm{2}} −\mathrm{6}{ap}−\mathrm{3}{ap}+\mathrm{9}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{p}\left({p}−\mathrm{3}{a}\right)−\mathrm{3}{a}\left({p}−\mathrm{3}{a}\right)=\mathrm{0} \\ $$$${p}=\frac{\mathrm{3}{a}}{\mathrm{2}},\:\:{or}\:\:{p}=\mathrm{3}{a}\:\left({i}\:{should}\:{reject}\right) \\ $$$${BC}={x}+{y}\:=\:{p}\sqrt{\mathrm{7}}\:=\:\left(\frac{\mathrm{3}\sqrt{\mathrm{7}}}{\mathrm{2}}\right){a} \\ $$$$\:\:\boldsymbol{{BC}}\:=\:\mathrm{150}\sqrt{\mathrm{7}}\:. \\ $$

Commented by mr W last updated on 10/Sep/19

thanks alot sir!

$${thanks}\:{alot}\:{sir}! \\ $$

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