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Question Number 208828 by efronzo1 last updated on 24/Jun/24
Answered by mr W last updated on 24/Jun/24
BM=MC=a, say AM=DM=2a ((BM)/6)=((MQ)/4) ⇒MQ=((2a)/3) ⇒QD=((4a)/3) (6+4)^2 =a^2 +(((2a)/3))^2 +a(((2a)/3)) ⇒a=((30)/( (√(19)))) ((sin ∠CBR)/((2a)/3))=((sin 120°)/(6+4)) sin ∠CBR=((√3)/(10×2))×(2/3)×((30)/( (√(19))))=((√3)/( (√(19)))) ((QR)/(sin 30°))=((QD)/(sin (90°+∠CBR)))=((4a)/(3 cos ∠CBR)) QR=((4×30)/(2×3(√(19))×(4/( (√(19))))))=5 ✓
$${BM}={MC}={a},\:{say} \\ $$$${AM}={DM}=\mathrm{2}{a} \\ $$$$\frac{{BM}}{\mathrm{6}}=\frac{{MQ}}{\mathrm{4}}\:\Rightarrow{MQ}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:\Rightarrow{QD}=\frac{\mathrm{4}{a}}{\mathrm{3}} \\ $$$$\left(\mathrm{6}+\mathrm{4}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\frac{\mathrm{2}{a}}{\mathrm{3}}\right)^{\mathrm{2}} +{a}\left(\frac{\mathrm{2}{a}}{\mathrm{3}}\right) \\ $$$$\Rightarrow{a}=\frac{\mathrm{30}}{\:\sqrt{\mathrm{19}}} \\ $$$$\frac{\mathrm{sin}\:\angle{CBR}}{\frac{\mathrm{2}{a}}{\mathrm{3}}}=\frac{\mathrm{sin}\:\mathrm{120}°}{\mathrm{6}+\mathrm{4}} \\ $$$$\mathrm{sin}\:\angle{CBR}=\frac{\sqrt{\mathrm{3}}}{\mathrm{10}×\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{30}}{\:\sqrt{\mathrm{19}}}=\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{19}}} \\ $$$$\frac{{QR}}{\mathrm{sin}\:\mathrm{30}°}=\frac{{QD}}{\mathrm{sin}\:\left(\mathrm{90}°+\angle{CBR}\right)}=\frac{\mathrm{4}{a}}{\mathrm{3}\:\mathrm{cos}\:\angle{CBR}} \\ $$$${QR}=\frac{\mathrm{4}×\mathrm{30}}{\mathrm{2}×\mathrm{3}\sqrt{\mathrm{19}}×\frac{\mathrm{4}}{\:\sqrt{\mathrm{19}}}}=\mathrm{5}\:\checkmark \\ $$
Commented by Tawa11 last updated on 24/Jun/24
Weldone sir. Learning.
$$\mathrm{Weldone}\:\mathrm{sir}.\:\mathrm{Learning}. \\ $$
Answered by A5T last updated on 24/Jun/24
Commented by A5T last updated on 24/Jun/24
((BM)/(MC))×((CD)/(DR))×((RQ)/(QB))=1⇒1×(t/(DR))×(?/(10))=1...(i) tan30°=((s/2)/t)⇒s=((2t(√3))/3);((sinθ)/(AP))=((sin30°)/6)⇒AP=12sinθ ((sin60°)/6)=((sin(90−θ))/(MP))⇒MP=4(√3)cosθ MP+AP=((2t(√3))/3)=12sinθ+4(√3)cosθ...(ii) ((sin(90−θ))/(CR))=((sinθ)/s)⇒CR=scotθ=((2t(√3)cotθ)/3) ⇒DR=((3t−2t(√3)cotθ)/3); sinθ=((2t(√3))/(30+3?)) ⇒30sinθ+3?sinθ=2t(√3)⇒?=((2t(√3)−30sinθ)/(3sinθ)) (i)⇒(3/(3−2(√3)cotθ))×((2t(√3)−30sinθ)/(30sinθ))=1 ⇒t(√3)=30sinθ−10(√3)cosθ...(iii) (ii)&(iii)⇒tanθ=((4(√3))/3)⇒sinθ=(4/( (√(19))))⇒cosθ=((√3)/( (√(19)))) ⇒t=((30(√3))/( (√(19))))⇒?=((2t(√3)−30sinθ)/(3sinθ))=5
$$\frac{{BM}}{{MC}}×\frac{{CD}}{{DR}}×\frac{{RQ}}{{QB}}=\mathrm{1}\Rightarrow\mathrm{1}×\frac{{t}}{{DR}}×\frac{?}{\mathrm{10}}=\mathrm{1}…\left({i}\right) \\ $$$${tan}\mathrm{30}°=\frac{\frac{{s}}{\mathrm{2}}}{{t}}\Rightarrow{s}=\frac{\mathrm{2}{t}\sqrt{\mathrm{3}}}{\mathrm{3}};\frac{{sin}\theta}{{AP}}=\frac{{sin}\mathrm{30}°}{\mathrm{6}}\Rightarrow{AP}=\mathrm{12}{sin}\theta \\ $$$$\frac{{sin}\mathrm{60}°}{\mathrm{6}}=\frac{{sin}\left(\mathrm{90}−\theta\right)}{{MP}}\Rightarrow{MP}=\mathrm{4}\sqrt{\mathrm{3}}{cos}\theta \\ $$$${MP}+{AP}=\frac{\mathrm{2}{t}\sqrt{\mathrm{3}}}{\mathrm{3}}=\mathrm{12}{sin}\theta+\mathrm{4}\sqrt{\mathrm{3}}{cos}\theta…\left({ii}\right) \\ $$$$\frac{{sin}\left(\mathrm{90}−\theta\right)}{{CR}}=\frac{{sin}\theta}{{s}}\Rightarrow{CR}={scot}\theta=\frac{\mathrm{2}{t}\sqrt{\mathrm{3}}{cot}\theta}{\mathrm{3}} \\ $$$$\Rightarrow{DR}=\frac{\mathrm{3}{t}−\mathrm{2}{t}\sqrt{\mathrm{3}}{cot}\theta}{\mathrm{3}};\:{sin}\theta=\frac{\mathrm{2}{t}\sqrt{\mathrm{3}}}{\mathrm{30}+\mathrm{3}?} \\ $$$$\Rightarrow\mathrm{30}{sin}\theta+\mathrm{3}?{sin}\theta=\mathrm{2}{t}\sqrt{\mathrm{3}}\Rightarrow?=\frac{\mathrm{2}{t}\sqrt{\mathrm{3}}−\mathrm{30}{sin}\theta}{\mathrm{3}{sin}\theta} \\ $$$$\left({i}\right)\Rightarrow\frac{\mathrm{3}}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}{cot}\theta}×\frac{\mathrm{2}{t}\sqrt{\mathrm{3}}−\mathrm{30}{sin}\theta}{\mathrm{30}{sin}\theta}=\mathrm{1} \\ $$$$\Rightarrow{t}\sqrt{\mathrm{3}}=\mathrm{30}{sin}\theta−\mathrm{10}\sqrt{\mathrm{3}}{cos}\theta…\left({iii}\right) \\ $$$$\left({ii}\right)\&\left({iii}\right)\Rightarrow{tan}\theta=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\Rightarrow{sin}\theta=\frac{\mathrm{4}}{\:\sqrt{\mathrm{19}}}\Rightarrow{cos}\theta=\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{19}}} \\ $$$$\Rightarrow{t}=\frac{\mathrm{30}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{19}}}\Rightarrow?=\frac{\mathrm{2}{t}\sqrt{\mathrm{3}}−\mathrm{30}{sin}\theta}{\mathrm{3}{sin}\theta}=\mathrm{5} \\ $$
Answered by A5T last updated on 24/Jun/24
1×((CD)/(DR))×(?/(10))=1..(i)[△BCR with transversal MQD] ((BM)/6)=((MQ)/4)⇒BM=3x,MQ=2x ((DR)/(RC))×((6x)/(3x))×((2x)/(4x))=1⇒DR=RC[△MCD,trns.BQR] (i)⇒2×(?/(10))=1⇒?=((10)/2)=5
$$\mathrm{1}×\frac{{CD}}{{DR}}×\frac{?}{\mathrm{10}}=\mathrm{1}..\left({i}\right)\left[\bigtriangleup{BCR}\:{with}\:{transversal}\:{MQD}\right] \\ $$$$\frac{{BM}}{\mathrm{6}}=\frac{{MQ}}{\mathrm{4}}\Rightarrow{BM}=\mathrm{3}{x},{MQ}=\mathrm{2}{x} \\ $$$$\frac{{DR}}{{RC}}×\frac{\mathrm{6}{x}}{\mathrm{3}{x}}×\frac{\mathrm{2}{x}}{\mathrm{4}{x}}=\mathrm{1}\Rightarrow{DR}={RC}\left[\bigtriangleup{MCD},{trns}.{BQR}\right] \\ $$$$\left({i}\right)\Rightarrow\mathrm{2}×\frac{?}{\mathrm{10}}=\mathrm{1}\Rightarrow?=\frac{\mathrm{10}}{\mathrm{2}}=\mathrm{5} \\ $$
Commented by A5T last updated on 24/Jun/24

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