Question-47354

Question Number 47354 by behi83417@gmail.com last updated on 08/Nov/18

Commented by behi83417@gmail.com last updated on 08/Nov/18

ABC,is a equelateral triangle with AB=2,and:AC ′=3,BA′=2. ......CD=?

$${ABC},{is}\:{a}\:{equelateral}\:{triangle}\:{with} \\ $$$${AB}=\mathrm{2},{and}:{AC}\:'=\mathrm{3},{BA}'=\mathrm{2}. \\ $$$$……\boldsymbol{\mathrm{CD}}=? \\ $$

Commented by ajfour last updated on 09/Nov/18

Commented by ajfour last updated on 09/Nov/18

γ = 30° ⇒ ∠C = 90° A′C =(√(4^2 −2^2 )) = 2(√3) ...(i) From △BCD ((sin β)/2)=((sin γ)/(BD)) & From △BDA′ ((sin γ)/(BD)) = ((sin (β−γ))/(A′D)) ⇒ A′D = ((3sin (β−γ))/(sin β)) ....(ii) From △CC ′D ((sin α)/(CD)) = ((cos β)/(CD))=((sin β)/5) ⇒ tan β = (5/(CD)) ...(iii) Using (i), (ii) for CD+A′D = A′C ⇒ CD+((3sin (β−γ))/(sin β)) = 2(√3) ...(I) let CD = x Then using (iii)&(I) x+3(cos γ−((sin γ)/(tan β)))=2(√3) ⇒ x+3(((√3)/2)−(x/(10)))=2(√3) or ((7x)/(10)) = ((√3)/2) ⇒ x = CD = ((5(√3))/7) .

$$\:\:\gamma\:=\:\mathrm{30}°\:\:\:\Rightarrow\:\angle{C}\:=\:\mathrm{90}° \\ $$$${A}'{C}\:=\sqrt{\mathrm{4}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }\:=\:\mathrm{2}\sqrt{\mathrm{3}}\:\:\:…\left({i}\right) \\ $$$${From}\:\bigtriangleup{BCD} \\ $$$$\:\frac{\mathrm{sin}\:\beta}{\mathrm{2}}=\frac{\mathrm{sin}\:\gamma}{{BD}}\:\:\&\:\:{From}\:\bigtriangleup{BDA}' \\ $$$$\:\frac{\mathrm{sin}\:\gamma}{{BD}}\:=\:\frac{\mathrm{sin}\:\left(\beta−\gamma\right)}{{A}'{D}} \\ $$$$\Rightarrow\:\:\:{A}'{D}\:=\:\frac{\mathrm{3sin}\:\left(\beta−\gamma\right)}{\mathrm{sin}\:\beta}\:\:\:….\left({ii}\right) \\ $$$${From}\:\bigtriangleup{CC}\:'{D} \\ $$$$\frac{\mathrm{sin}\:\alpha}{{CD}}\:=\:\frac{\mathrm{cos}\:\beta}{{CD}}=\frac{\mathrm{sin}\:\beta}{\mathrm{5}}\: \\ $$$$\Rightarrow\:\:\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{5}}{{CD}}\:\:\:\:…\left({iii}\right) \\ $$$${Using}\:\:\left({i}\right),\:\left({ii}\right)\:{for} \\ $$$$\:\:\:{CD}+{A}'{D}\:=\:{A}'{C} \\ $$$$\Rightarrow\:\:{CD}+\frac{\mathrm{3sin}\:\left(\beta−\gamma\right)}{\mathrm{sin}\:\beta}\:=\:\mathrm{2}\sqrt{\mathrm{3}}\:\:\:…\left({I}\right) \\ $$$${let}\:\:{CD}\:=\:{x} \\ $$$${Then}\:{using}\:\left({iii}\right)\&\left({I}\right) \\ $$$$\:\:\:\:{x}+\mathrm{3}\left(\mathrm{cos}\:\gamma−\frac{\mathrm{sin}\:\gamma}{\mathrm{tan}\:\beta}\right)=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:\:{x}+\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{{x}}{\mathrm{10}}\right)=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${or}\:\:\:\:\:\frac{\mathrm{7}{x}}{\mathrm{10}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:{x}\:=\:{CD}\:=\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{7}}\:.\: \\ $$

Commented by behi83417@gmail.com last updated on 09/Nov/18

$${sir}\:{Ajfour}!{thank}\:{you}\:{very}\:{much} \\ $$$${for}\:{nice}\:{work}.{god}\:{bless}\:{you}\:{sir}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply