Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com