Question-182608

Question Number 182608 by Acem last updated on 11/Dec/22

Commented by Acem last updated on 11/Dec/22

$${Find}\:{the}\:{tangent}\:{lenght}\:{of}\:{the}\:{two}\:{small}\:{circles} \\ $$$$\:{it}\:{intersects}\:{with}\:{the}\:{big}\:{circle}\:{in}\:{two}\:{points} \\ $$

Answered by mr W last updated on 11/Dec/22

Commented by mr W last updated on 12/Dec/22

cos α=((7−2)/(7+2))=(5/9) cos (π−β−α)=((3^2 +9^2 −8^2 )/(2×3×9))=((13)/(27)) β=π−cos^(−1) ((13)/(27))−cos^(−1) (5/9) cos β=−cos (cos^(−1) ((13)/(27))+cos^(−1) (5/9)) =((4(√(35)))/(27))×((2(√(14)))/9)−((13)/(27))×(5/9)=((56(√(10))−65)/(243)) OE=7+3 cos β =7+3×((56(√(10))−65)/(243))=((56(√(10))+502)/(81)) x=FG=2×(√(10^2 −(((56(√(10))+502)/(81)))^2 )) =((8(√(23296−3514(√(10)))))/(81))≈10.902

$$\mathrm{cos}\:\alpha=\frac{\mathrm{7}−\mathrm{2}}{\mathrm{7}+\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{9}} \\ $$$$\mathrm{cos}\:\left(\pi−\beta−\alpha\right)=\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\mathrm{9}}=\frac{\mathrm{13}}{\mathrm{27}} \\ $$$$\beta=\pi−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{13}}{\mathrm{27}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{9}} \\ $$$$\mathrm{cos}\:\beta=−\mathrm{cos}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{13}}{\mathrm{27}}+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{9}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}\sqrt{\mathrm{35}}}{\mathrm{27}}×\frac{\mathrm{2}\sqrt{\mathrm{14}}}{\mathrm{9}}−\frac{\mathrm{13}}{\mathrm{27}}×\frac{\mathrm{5}}{\mathrm{9}}=\frac{\mathrm{56}\sqrt{\mathrm{10}}−\mathrm{65}}{\mathrm{243}} \\ $$$${OE}=\mathrm{7}+\mathrm{3}\:\mathrm{cos}\:\beta \\ $$$$\:\:\:\:\:\:\:=\mathrm{7}+\mathrm{3}×\frac{\mathrm{56}\sqrt{\mathrm{10}}−\mathrm{65}}{\mathrm{243}}=\frac{\mathrm{56}\sqrt{\mathrm{10}}+\mathrm{502}}{\mathrm{81}} \\ $$$${x}={FG}=\mathrm{2}×\sqrt{\mathrm{10}^{\mathrm{2}} −\left(\frac{\mathrm{56}\sqrt{\mathrm{10}}+\mathrm{502}}{\mathrm{81}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{8}\sqrt{\mathrm{23296}−\mathrm{3514}\sqrt{\mathrm{10}}}}{\mathrm{81}}\approx\mathrm{10}.\mathrm{902} \\ $$

Commented by Acem last updated on 12/Dec/22

Great Sir! Please just replace α instead γ

$${Great}\:{Sir}!\: \\ $$$$\:{Please}\:{just}\:{replace}\:\alpha\:{instead}\:\gamma\: \\ $$

Commented by mr W last updated on 12/Dec/22

the other chord: cos (α−γ)=((3^2 +9^2 −8^2 )/(2×3×9))=((13)/(27)) γ=cos^(−1) (5/9)−cos^(−1) ((13)/(27)) cos γ=((56(√(10))+65)/(243)) OE′=7−3 cos γ =7−3×((56(√(10))+65)/(243))=((502−56(√(10)))/(81)) x=2×(√(10^2 −(((−56(√(10))+502)/(81)))^2 )) =((8(√(23296+3514(√(10)))))/(81))≈18.320

$${the}\:{other}\:{chord}: \\ $$$$\mathrm{cos}\:\left(\alpha−\gamma\right)=\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\mathrm{9}}=\frac{\mathrm{13}}{\mathrm{27}} \\ $$$$\gamma=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{9}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{13}}{\mathrm{27}} \\ $$$$\mathrm{cos}\:\gamma=\frac{\mathrm{56}\sqrt{\mathrm{10}}+\mathrm{65}}{\mathrm{243}} \\ $$$${OE}'=\mathrm{7}−\mathrm{3}\:\mathrm{cos}\:\gamma \\ $$$$\:\:\:\:\:\:\:=\mathrm{7}−\mathrm{3}×\frac{\mathrm{56}\sqrt{\mathrm{10}}+\mathrm{65}}{\mathrm{243}}=\frac{\mathrm{502}−\mathrm{56}\sqrt{\mathrm{10}}}{\mathrm{81}} \\ $$$${x}=\mathrm{2}×\sqrt{\mathrm{10}^{\mathrm{2}} −\left(\frac{−\mathrm{56}\sqrt{\mathrm{10}}+\mathrm{502}}{\mathrm{81}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{8}\sqrt{\mathrm{23296}+\mathrm{3514}\sqrt{\mathrm{10}}}}{\mathrm{81}}\approx\mathrm{18}.\mathrm{320} \\ $$

Commented by mr W last updated on 12/Dec/22

Commented by Acem last updated on 12/Dec/22

$${Oh}\:{why}\:{not}!\:{I}\:{will}\:{note}\:{it}\:{in}\:{my}\:{book}, \\ $$$$\:{Thank}\:{you}\:{very}\:{much}\:{Sir}! \\ $$

Commented by Acem last updated on 12/Dec/22

I found another way to solve it, but it′s very annoying and boring (√(39− 10y −y^2 )) + (√(9−y^2 ))= 2(√(14)) ; y is the same dis. 3 cos β _(For the 1st chord) So, and on here using cosine laws is better than Pythagoras one Here cosine laws are as the magical solution!

$${I}\:{found}\:{another}\:{way}\:{to}\:{solve}\:{it},\:{but}\:{it}'{s}\:{very} \\ $$$$\:{annoying}\:{and}\:{boring} \\ $$$$\:\sqrt{\mathrm{39}−\:\mathrm{10}{y}\:−{y}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{9}−{y}^{\mathrm{2}} }=\:\mathrm{2}\sqrt{\mathrm{14}}\:\: \\ $$$$\:;\:{y}\:{is}\:{the}\:{same}\:{dis}.\:\mathrm{3}\:\mathrm{cos}\:\beta\:_{{For}\:{the}\:\mathrm{1}{st}\:{chord}} \\ $$$$\:{So},\:{and}\:{on}\:{here}\:{using}\:{cosine}\:{laws}\:{is}\:{better} \\ $$$$\:{than}\:{Pythagoras}\:{one} \\ $$$$\:{Here}\:{cosine}\:{laws}\:{are}\:{as}\:{the}\:{magical}\:{solution}! \\ $$$$\: \\ $$

Answered by Acem last updated on 12/Dec/22

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