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Question Number 147259 by mathdanisur last updated on 19/Jul/21

Answered by Rasheed.Sindhi last updated on 19/Jul/21

△AED: ∵AD is diameter ∴∠AED=90 tan∠A=(1/3) ⇒∠A=19.47 AC=(√(AE^2 +EC^2 ))=(√(3^2 +1^2 ))=(√(10)) △ABE: AB=AC=(√(10)) ∠BAE=90+19.47=109.47 EB^2 =AB^2 +AE^2 −2AB.AE.cos∠BAE EB^2 =((√(10)))^2 +(3)^2 −2(√(10)).3.cos 109.47 =10+9−6(√(10)) (−(1/3)) =19+2(√(10)) EB=(√(19+2(√(10))))≈5.03

$$\bigtriangleup{AED}: \\ $$$$\because{AD}\:{is}\:{diameter} \\ $$$$\therefore\angle{AED}=\mathrm{90} \\ $$$${tan}\angle{A}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\angle{A}=\mathrm{19}.\mathrm{47} \\ $$$${AC}=\sqrt{{AE}^{\mathrm{2}} +{EC}^{\mathrm{2}} }=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\sqrt{\mathrm{10}} \\ $$$$\bigtriangleup{ABE}: \\ $$$${AB}={AC}=\sqrt{\mathrm{10}} \\ $$$$\angle{BAE}=\mathrm{90}+\mathrm{19}.\mathrm{47}=\mathrm{109}.\mathrm{47} \\ $$$${EB}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{AE}^{\mathrm{2}} −\mathrm{2}{AB}.{AE}.{cos}\angle{BAE} \\ $$$${EB}^{\mathrm{2}} =\left(\sqrt{\mathrm{10}}\right)^{\mathrm{2}} +\left(\mathrm{3}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{10}}.\mathrm{3}.{cos}\:\mathrm{109}.\mathrm{47} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{10}+\mathrm{9}−\mathrm{6}\sqrt{\mathrm{10}}\:\left(−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{19}+\mathrm{2}\sqrt{\mathrm{10}}\: \\ $$$${EB}=\sqrt{\mathrm{19}+\mathrm{2}\sqrt{\mathrm{10}}}\approx\mathrm{5}.\mathrm{03} \\ $$

Commented by mathdanisur last updated on 19/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

Commented by otchereabdullai@gmail.com last updated on 19/Jul/21

nice!

$$\mathrm{nice}! \\ $$

Answered by liberty last updated on 19/Jul/21

AD=(√(3^2 +1^2 )) =(√(10)) = AB let ∠EAD = α cos α = ((9+10−1)/(2.3.(√(10)))) = (3/( (√(10)))) then sin α= (1/( (√(10)))) ∠EAB = 90°+α cos (90°+α)=((9+10−EB^2 )/(2.3.(√(10)))) ⇒−sin α = ((19−EB^2 )/(6(√(10)))) ⇒−(1/( (√(10))))×6(√(10)) = 19−EB^2 ⇒EB =(√(19+6)) = 5

$${AD}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }\:=\sqrt{\mathrm{10}}\:=\:{AB} \\ $$$${let}\:\angle{EAD}\:=\:\alpha \\ $$$$\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{9}+\mathrm{10}−\mathrm{1}}{\mathrm{2}.\mathrm{3}.\sqrt{\mathrm{10}}}\:=\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}\:{then} \\ $$$$\mathrm{sin}\:\alpha=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}} \\ $$$$\angle{EAB}\:=\:\mathrm{90}°+\alpha \\ $$$$\mathrm{cos}\:\left(\mathrm{90}°+\alpha\right)=\frac{\mathrm{9}+\mathrm{10}−{EB}^{\mathrm{2}} }{\mathrm{2}.\mathrm{3}.\sqrt{\mathrm{10}}} \\ $$$$\Rightarrow−\mathrm{sin}\:\alpha\:=\:\frac{\mathrm{19}−{EB}^{\mathrm{2}} }{\mathrm{6}\sqrt{\mathrm{10}}} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}×\mathrm{6}\sqrt{\mathrm{10}}\:=\:\mathrm{19}−{EB}^{\mathrm{2}} \\ $$$$\Rightarrow{EB}\:=\sqrt{\mathrm{19}+\mathrm{6}}\:=\:\mathrm{5}\: \\ $$

Commented by mathdanisur last updated on 19/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

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