Question Number 222921 by cherokeesay last updated on 10/Jul/25
Answered by mr W last updated on 11/Jul/25
Commented by mr W last updated on 11/Jul/25
$$\frac{{k}}{\mathrm{2}{k}}×\frac{\mathrm{1}}{\mathrm{3}}×\frac{{p}+{q}}{{q}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{p}}{{q}}=\mathrm{5} \\ $$$$\frac{{p}}{{q}}×\frac{{x}}{{y}}×\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{0}.\mathrm{8}\:\checkmark \\ $$
Commented by mr W last updated on 11/Jul/25
Commented by fantastic last updated on 11/Jul/25
$${Sir}\:{how}\:{do}\:{you}\:{come}\:{to}\:{know}\:{about}\:{these}\:{properties}? \\ $$$${Can}\:{you}\:{please}\:{inform}\:{me}\:{about}\:{more}\:{interesting}\:{properties}? \\ $$
Commented by mr W last updated on 11/Jul/25
Commented by mr W last updated on 11/Jul/25
$$\frac{{p}}{{q}}×\frac{\mathrm{2}{k}}{\mathrm{1}{k}}×\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{p}}{{q}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{x}}{{y}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{{p}+{q}}{{q}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{4}}{\mathrm{5}}\:\checkmark \\ $$
Commented by cherokeesay last updated on 11/Jul/25
$${thank}\:{you}\:{sir}. \\ $$
Commented by fantastic last updated on 11/Jul/25
$$\mathrm{v}.\cap{i}\mathbb{C}\epsilon! \\ $$
Commented by mr W last updated on 11/Jul/25
Answered by fantastic last updated on 11/Jul/25
Commented by fantastic last updated on 11/Jul/25
![Side of eqi. traingle=3+1=4 AR ,BP,CQ are medians So PC=(4/2)=2 ∡PCG=(1/2)×60^0 =30^0 amd ∡BPC=90^0 ((GP)/(PC))=tan 30^0 =(1/( (√3))) GP=((PC)/( (√3)))=(2/( (√3))) and (2/(GC))=cos 30^0 =((√3)/2) GC=((2×2)/( (√3)))=(4/( (√3))) PB=1 and GP=(2/( (√3))) if ∡GBP=θ then θ=tan^(−1) (((2/( (√3)))/1))=tan^(−1) ((2/( (√3)))) So ∡CGB=(tan^(−1) ((2/( (√3))))−30^0 ) So ∡EGQ=(tan^(−1) ((2/( (√3))))−30^0 )=α[let] Now CQ=((√3)/2)×4=2(√3) and CG=(4/( (√3))) So GQ=(2(√3)−(4/( (√3))))=2((√3)−(2/( (√3))))=2(((3−2)/( (√3))))=(2/( (√3))) So ((GQ)/(GE))=cos α GE=((2/( (√3)))/(cos α))=(2/(cos α(√3))) Now ((GP)/(GB))=sin θ GB=((GP)/(sin θ))=((2/( (√3)))/(sin θ))=(2/(sin θ(√3))) SO ((GE)/(GB))=((2/(cos α(√3)))/(2/(sin θ(√3))))=(2/(cos α(√3)))×((sin θ(√3))/2)=((sin θ)/(cos α)) ((sin (tan^(−1) ((2/( (√3))))))/(cos (tan^(−1) ((2/( (√3))))−30^0 )))=0.8 determinant (((Final Answer ((GE)/(GB))=0.8)))✓✓](https://www.tinkutara.com/question/Q222936.png)
$${Side}\:{of}\:{eqi}.\:{traingle}=\mathrm{3}+\mathrm{1}=\mathrm{4} \\ $$$${AR}\:,{BP},{CQ}\:{are}\:{medians} \\ $$$${So}\:{PC}=\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{2} \\ $$$$\measuredangle{PCG}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{60}^{\mathrm{0}} =\mathrm{30}^{\mathrm{0}} \:{amd}\:\measuredangle{BPC}=\mathrm{90}^{\mathrm{0}} \\ $$$$\frac{{GP}}{{PC}}=\mathrm{tan}\:\mathrm{30}^{\mathrm{0}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${GP}=\frac{{PC}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$${and}\:\:\:\frac{\mathrm{2}}{{GC}}=\mathrm{cos}\:\mathrm{30}^{\mathrm{0}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${GC}=\frac{\mathrm{2}×\mathrm{2}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$${PB}=\mathrm{1}\:{and}\:{GP}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\: \\ $$$${if}\:\measuredangle{GBP}=\theta\:{then}\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${So}\:\measuredangle{CGB}=\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{30}^{\mathrm{0}} \right) \\ $$$${So}\:\measuredangle{EGQ}=\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{30}^{\mathrm{0}} \right)=\alpha\left[{let}\right] \\ $$$${Now}\:{CQ}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{4}=\mathrm{2}\sqrt{\mathrm{3}}\:{and}\:{CG}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\: \\ $$$${So}\:{GQ}=\left(\mathrm{2}\sqrt{\mathrm{3}}−\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{2}\left(\sqrt{\mathrm{3}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{2}\left(\frac{\mathrm{3}−\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$${So}\:\frac{{GQ}}{{GE}}=\mathrm{cos}\:\alpha \\ $$$${GE}=\frac{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}}{\mathrm{cos}\:\alpha}=\frac{\mathrm{2}}{\mathrm{cos}\:\alpha\sqrt{\mathrm{3}}} \\ $$$${Now}\:\frac{{GP}}{{GB}}=\mathrm{sin}\:\theta \\ $$$${GB}=\frac{{GP}}{\mathrm{sin}\:\theta}=\frac{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}}{\mathrm{sin}\:\theta}=\frac{\mathrm{2}}{\mathrm{sin}\:\theta\sqrt{\mathrm{3}}} \\ $$$${SO}\:\frac{{GE}}{{GB}}=\frac{\frac{\mathrm{2}}{\mathrm{cos}\:\alpha\sqrt{\mathrm{3}}}}{\frac{\mathrm{2}}{\mathrm{sin}\:\theta\sqrt{\mathrm{3}}}}=\frac{\cancel{\mathrm{2}}}{\mathrm{cos}\:\alpha\cancel{\sqrt{\mathrm{3}}}}×\frac{\mathrm{sin}\:\theta\cancel{\sqrt{\mathrm{3}}}}{\cancel{\mathrm{2}}}=\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\alpha} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)\right)}{\mathrm{cos}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{30}^{\mathrm{0}} \right)}=\mathrm{0}.\mathrm{8} \\ $$$$\begin{array}{|c|}{{Final}\:{Answer}\:\frac{{GE}}{{GB}}=\mathrm{0}.\mathrm{8}}\\\hline\end{array}\checkmark\checkmark \\ $$
Commented by mr W last updated on 11/Jul/25
$${the}\:{triangle}\:{mustn}'{t}\:{be}\:{equilateral}. \\ $$