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Question Number 62289 by naka3546 last updated on 19/Jun/19

Answered by MJS last updated on 19/Jun/19

we have A_1 =3=(1/2)×a×x ⇒ x=(6/a) A_2 =5=(1/2)×b×y ⇒ y=((10)/b) side of triangle = c (1) a^2 +((36)/a^2 )=c^2 (2) b^2 +((100)/b^2 )=c^2 (3) (a−((10)/b))^2 +(b−(6/a))^2 =c^2 solving these leads to a=((972))^(1/4) b=(((2500)/3))^(1/4) c=(((3136)/3))^(1/4) or a=((4/3))^(1/4) b=((12))^(1/4) c=(((3136)/3))^(1/4) ⇒ A_3 =8

$$\mathrm{we}\:\mathrm{have} \\ $$$${A}_{\mathrm{1}} =\mathrm{3}=\frac{\mathrm{1}}{\mathrm{2}}×{a}×{x}\:\Rightarrow\:{x}=\frac{\mathrm{6}}{{a}} \\ $$$${A}_{\mathrm{2}} =\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}×{b}×{y}\:\Rightarrow\:{y}=\frac{\mathrm{10}}{{b}} \\ $$$$\mathrm{side}\:\mathrm{of}\:\mathrm{triangle}\:=\:{c} \\ $$$$\left(\mathrm{1}\right)\:{a}^{\mathrm{2}} +\frac{\mathrm{36}}{{a}^{\mathrm{2}} }={c}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{b}^{\mathrm{2}} +\frac{\mathrm{100}}{{b}^{\mathrm{2}} }={c}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\left({a}−\frac{\mathrm{10}}{{b}}\right)^{\mathrm{2}} +\left({b}−\frac{\mathrm{6}}{{a}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\mathrm{solving}\:\mathrm{these}\:\mathrm{leads}\:\mathrm{to} \\ $$$${a}=\sqrt[{\mathrm{4}}]{\mathrm{972}}\:\:{b}=\sqrt[{\mathrm{4}}]{\frac{\mathrm{2500}}{\mathrm{3}}}\:\:{c}=\sqrt[{\mathrm{4}}]{\frac{\mathrm{3136}}{\mathrm{3}}} \\ $$$$\mathrm{or} \\ $$$${a}=\sqrt[{\mathrm{4}}]{\frac{\mathrm{4}}{\mathrm{3}}}\:\:{b}=\sqrt[{\mathrm{4}}]{\mathrm{12}}\:\:{c}=\sqrt[{\mathrm{4}}]{\frac{\mathrm{3136}}{\mathrm{3}}} \\ $$$$\Rightarrow\:{A}_{\mathrm{3}} =\mathrm{8} \\ $$

Commented by MJS last updated on 19/Jun/19

btw only the 1^(st) solution looks like in the picture

$$\mathrm{btw}\:\mathrm{only}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{solution}\:\mathrm{looks}\:\mathrm{like}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{picture} \\ $$

Commented by mr W last updated on 19/Jun/19

exact solution, very nice sir! i think A_3 =A_1 +A_2 , proof see below.

$${exact}\:{solution},\:{very}\:{nice}\:{sir}! \\ $$$${i}\:{think}\:{A}_{\mathrm{3}} ={A}_{\mathrm{1}} +{A}_{\mathrm{2}} ,\:{proof}\:{see}\:{below}. \\ $$

Answered by mr W last updated on 19/Jun/19

Commented by mr W last updated on 19/Jun/19

ϕ=(π/2)−(π/3)−θ=(π/6)−θ φ=π−((π/2)−θ)−(π/3)=(π/6)+θ A_1 =((c^2 sin 2θ)/4) A_2 =((c^2 sin 2ϕ)/4)=((c^2 sin ((π/3)−2θ))/4) A_3 =((c^2 sin 2φ)/4)=((c^2 sin ((π/3)+2θ))/4) (A_2 /A_1 )=((sin ((π/3)−2θ))/(sin 2θ)) ((((√3)/2) cos 2θ−(1/2)sin 2θ)/(sin 2θ))=(A_2 /A_1 ) ⇒((√3)/(tan 2θ))=((2A_2 )/A_1 )+1 (A_3 /A_1 )=((sin ((π/3)+2θ))/(sin 2θ)) =(1/2)(((√3)/(tan 2θ))+1)=(1/2)(((2A_2 )/A_1 )+1+1)=1+(A_2 /A_1 ) ⇒A_3 =A_1 (1+(A_2 /A_1 ))=A_1 +A_2 =3+5=8

$$\varphi=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}−\theta=\frac{\pi}{\mathrm{6}}−\theta \\ $$$$\phi=\pi−\left(\frac{\pi}{\mathrm{2}}−\theta\right)−\frac{\pi}{\mathrm{3}}=\frac{\pi}{\mathrm{6}}+\theta \\ $$$${A}_{\mathrm{1}} =\frac{{c}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{4}} \\ $$$${A}_{\mathrm{2}} =\frac{{c}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\varphi}{\mathrm{4}}=\frac{{c}^{\mathrm{2}} \:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{2}\theta\right)}{\mathrm{4}} \\ $$$${A}_{\mathrm{3}} =\frac{{c}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{4}}=\frac{{c}^{\mathrm{2}} \:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\mathrm{2}\theta\right)}{\mathrm{4}} \\ $$$$\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{2}\theta\right)}{\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} } \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{tan}\:\:\mathrm{2}\theta}=\frac{\mathrm{2}{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }+\mathrm{1} \\ $$$$\frac{{A}_{\mathrm{3}} }{{A}_{\mathrm{1}} }=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\mathrm{2}\theta\right)}{\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{tan}\:\mathrm{2}\theta}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }+\mathrm{1}+\mathrm{1}\right)=\mathrm{1}+\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} } \\ $$$$\Rightarrow{A}_{\mathrm{3}} ={A}_{\mathrm{1}} \left(\mathrm{1}+\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }\right)={A}_{\mathrm{1}} +{A}_{\mathrm{2}} =\mathrm{3}+\mathrm{5}=\mathrm{8} \\ $$

Commented by Tony Lin last updated on 20/Jun/19

2(A_1 +A_2 )=c^2 sinθcosθ+c^2 sin((π/6)−θ)cos((π/6)−θ) =(1/2)c^2 [sin2θ+sin((π/3)−2θ)] =(1/2)c^2 (((√3)/2)cos2θ+(1/2)sin2θ) =(1/2)c^2 sin((π/3)+2θ) =c^2 cos((π/6)+θ)sin((π/6)+θ) =2A_3

$$\mathrm{2}\left({A}_{\mathrm{1}} +{A}_{\mathrm{2}} \right)={c}^{\mathrm{2}} {sin}\theta{cos}\theta+{c}^{\mathrm{2}} {sin}\left(\frac{\pi}{\mathrm{6}}−\theta\right){cos}\left(\frac{\pi}{\mathrm{6}}−\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{c}^{\mathrm{2}} \left[{sin}\mathrm{2}\theta+{sin}\left(\frac{\pi}{\mathrm{3}}−\mathrm{2}\theta\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{c}^{\mathrm{2}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{c}^{\mathrm{2}} {sin}\left(\frac{\pi}{\mathrm{3}}+\mathrm{2}\theta\right) \\ $$$$={c}^{\mathrm{2}} {cos}\left(\frac{\pi}{\mathrm{6}}+\theta\right){sin}\left(\frac{\pi}{\mathrm{6}}+\theta\right) \\ $$$$=\mathrm{2}{A}_{\mathrm{3}} \\ $$

Answered by ajfour last updated on 19/Jun/19

c^2 sin θcos θ=6 c^2 sin φcos φ=10 ⇒ ((sin 2θ)/(sin 2φ))=(3/5) θ+φ=(π/6) ⇒ 2θ+2φ=(π/3) ⇒ 5sin 2θ= 3sin ((π/3)−2θ) ⇒ 10sin 2θ=3(√3)cos 2θ−3sin 2θ ⇒ tan 2θ=((3(√3))/(13)) ⇒ cos 2θ=((13)/(14)) & tan 2φ=tan ((π/3)−2θ) =(((√3)−((3(√3))/(13)))/(1+(9/(13)))) = ((5(√3))/(11)) ⇒ cos 2φ=((11)/(14)) cos 2θ+cos 2φ=2cos (θ+φ)cos (θ−φ) ⇒ ((12)/7)= (√3)cos (θ−φ) ⇒ cos (θ−φ)=((12)/(7(√3))) c^2 =((12)/(sin 2θ)) = ((12)/((3(√3)/14)))=((56)/(√3)) c^2 (cos θ−sin φ)(cos φ−sin θ)= 2A_3 ⇒ 2A_3 = c^2 (cos θcos φ+sin θsin φ) −c^2 (cos θsin θ+sin φcos φ) ⇒ 2A_3 =c^2 cos (θ−φ)−16 = ((56)/(√3))×((12)/(7(√3)))−16 = 32−16 ⇒ A_3 = 8 .

$${c}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta=\mathrm{6} \\ $$$${c}^{\mathrm{2}} \mathrm{sin}\:\phi\mathrm{cos}\:\phi=\mathrm{10}\:\:\:\Rightarrow\:\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{sin}\:\mathrm{2}\phi}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\theta+\phi=\frac{\pi}{\mathrm{6}}\:\:\:\:\:\Rightarrow\:\:\mathrm{2}\theta+\mathrm{2}\phi=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:\mathrm{5sin}\:\mathrm{2}\theta=\:\mathrm{3sin}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{2}\theta\right) \\ $$$$\Rightarrow\:\:\mathrm{10sin}\:\mathrm{2}\theta=\mathrm{3}\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta−\mathrm{3sin}\:\mathrm{2}\theta \\ $$$$\Rightarrow\:\mathrm{tan}\:\mathrm{2}\theta=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{13}}\:\:\Rightarrow\:\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{13}}{\mathrm{14}} \\ $$$$\&\:\:\mathrm{tan}\:\mathrm{2}\phi=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{2}\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{13}}}{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{13}}}\:=\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{11}} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\mathrm{2}\phi=\frac{\mathrm{11}}{\mathrm{14}} \\ $$$$\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\mathrm{2}\phi=\mathrm{2cos}\:\left(\theta+\phi\right)\mathrm{cos}\:\left(\theta−\phi\right) \\ $$$$\Rightarrow\:\frac{\mathrm{12}}{\mathrm{7}}=\:\sqrt{\mathrm{3}}\mathrm{cos}\:\left(\theta−\phi\right) \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\left(\theta−\phi\right)=\frac{\mathrm{12}}{\mathrm{7}\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:{c}^{\mathrm{2}} =\frac{\mathrm{12}}{\mathrm{sin}\:\mathrm{2}\theta}\:=\:\frac{\mathrm{12}}{\left(\mathrm{3}\sqrt{\mathrm{3}}/\mathrm{14}\right)}=\frac{\mathrm{56}}{\sqrt{\mathrm{3}}} \\ $$$${c}^{\mathrm{2}} \left(\mathrm{cos}\:\theta−\mathrm{sin}\:\phi\right)\left(\mathrm{cos}\:\phi−\mathrm{sin}\:\theta\right)=\:\mathrm{2}{A}_{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{2}{A}_{\mathrm{3}} =\:{c}^{\mathrm{2}} \left(\mathrm{cos}\:\theta\mathrm{cos}\:\phi+\mathrm{sin}\:\theta\mathrm{sin}\:\phi\right) \\ $$$$\:\:\:\:\:\:\:\:−{c}^{\mathrm{2}} \left(\mathrm{cos}\:\theta\mathrm{sin}\:\theta+\mathrm{sin}\:\phi\mathrm{cos}\:\phi\right) \\ $$$$\Rightarrow\:\mathrm{2}{A}_{\mathrm{3}} ={c}^{\mathrm{2}} \mathrm{cos}\:\left(\theta−\phi\right)−\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{56}}{\sqrt{\mathrm{3}}}×\frac{\mathrm{12}}{\mathrm{7}\sqrt{\mathrm{3}}}−\mathrm{16}\:=\:\mathrm{32}−\mathrm{16} \\ $$$$\Rightarrow\:\:\boldsymbol{{A}}_{\mathrm{3}} =\:\mathrm{8}\:. \\ $$

Commented by ajfour last updated on 19/Jun/19

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