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Question Number 220769 by hardmath last updated on 18/May/25
Answered by mr W last updated on 18/May/25
Commented by mr W last updated on 18/May/25
(x/( (√(a^2 +b^2 ))))=(c/a)=((√(a^2 +b^2 ))/b) ⇒x=((a^2 +b^2 )/b) ⇒c=((a(√(a^2 +b^2 )))/b) y=(√(c^2 −b^2 ))=(√(((a^2 (a^2 +b^2 ))/b^2 )−b^2 )) (2a)^2 +x^2 =(a+y)^2 3a^2 +x^2 =2ay+y^2 3a^2 +(((a^2 +b^2 )^2 )/b^2 )=2a(√(((a^2 (a^2 +b^2 ))/b^2 )−b^2 ))+((a^2 (a^2 +b^2 ))/b^2 )−b^2 2a^2 +b^2 =a(√(((a^2 (a^2 +b^2 ))/b^2 )−b^2 )) 4a^4 +4a^2 b^2 +b^4 =((a^4 (a^2 +b^2 ))/b^2 )−a^2 b^2 (4a^2 +b^2 )(a^2 +b^2 )=((a^4 (a^2 +b^2 ))/b^2 ) 4a^2 +b^2 =(a^4 /b^2 ) ((a/b))^4 −4((a/b))^2 −1=0 ⇒((a/b))^2 =2+(√5) ((CD)/(AB))=(c/(2a))=((a(√(a^2 +b^2 )))/(2ab))=(1/2)(√(1+((a/b))^2 )) =((√(3+(√5)))/2)=(((√2)+(√(10)))/4)≈1.144
$$\frac{{x}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}=\frac{{c}}{{a}}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{b}} \\ $$$$\Rightarrow{x}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}} \\ $$$$\Rightarrow{c}=\frac{{a}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{b}} \\ $$$${y}=\sqrt{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\sqrt{\frac{{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} }−{b}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}{a}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} =\left({a}+{y}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{2}{ay}+{y}^{\mathrm{2}} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} +\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{2}{a}\sqrt{\frac{{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} }−{b}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} }−{b}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}\sqrt{\frac{{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} }−{b}^{\mathrm{2}} } \\ $$$$\mathrm{4}{a}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} =\frac{{a}^{\mathrm{4}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} }−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left(\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\frac{{a}^{\mathrm{4}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} } \\ $$$$\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{2}} } \\ $$$$\left(\frac{{a}}{{b}}\right)^{\mathrm{4}} −\mathrm{4}\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\frac{{CD}}{{AB}}=\frac{{c}}{\mathrm{2}{a}}=\frac{{a}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}{ab}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{10}}}{\mathrm{4}}\approx\mathrm{1}.\mathrm{144} \\ $$
Commented by Tawa11 last updated on 18/May/25
Weldone sir. Sir, I asked question on your proposed question Q216060
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$$$\mathrm{Sir},\:\mathrm{I}\:\mathrm{asked}\:\mathrm{question}\:\mathrm{on}\:\mathrm{your}\:\mathrm{proposed}\:\mathrm{question} \\ $$$$\mathrm{Q216060} \\ $$
Answered by mr W last updated on 19/May/25
Commented by mr W last updated on 19/May/25
acc. to law of sines: (1/(tan α sin α))=(1/(sin ((π/2)−3α))) ((cos α)/(sin^2 α))=(1/(cos α(4 cos^2 α−3))) 4 cos^4 α−3 cos^2 α=1−cos^2 α 4 cos^4 α−2 cos^2 α−1=0 cos^2 α=((1+(√5))/4) ⇒sin α=(√(1−((1+(√5))/4)))=(((√(10))−(√2))/4) ((CD)/(AB))=(1/(2 cos α tan α ))=(1/(2 sin α)) =(2/( (√(10))−(√2)))=(((√(10))+(√2))/4) ✓
$${acc}.\:{to}\:{law}\:{of}\:{sines}: \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\alpha\:\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{3}\alpha\right)} \\ $$$$\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}^{\mathrm{2}} \:\alpha}=\frac{\mathrm{1}}{\mathrm{cos}\:\alpha\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{3}\right)} \\ $$$$\mathrm{4}\:\mathrm{cos}^{\mathrm{4}} \:\alpha−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\alpha=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha \\ $$$$\mathrm{4}\:\mathrm{cos}^{\mathrm{4}} \:\alpha−\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\alpha=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\: \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\sqrt{\mathrm{1}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{10}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\frac{{CD}}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\alpha\:\mathrm{tan}\:\alpha\:}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\alpha} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{10}}−\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{10}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\:\checkmark \\ $$
Commented by hardmath last updated on 19/May/25
cool my dear professor thank you very much
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

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