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Question-179088




Question Number 179088 by cortano1 last updated on 24/Oct/22
Answered by mr W last updated on 24/Oct/22
Method 1  cos α=(7/R)  R^2 −7^2 =58^2 +92^2 +2×58×92×(7/R)  R^3 −11877R−71704=0  R=2(√(3959)) sin (((2π)/3)−(1/3)sin^(−1) ((35852)/(3959(√(3959)))))      ≈112.095
$$\boldsymbol{{Method}}\:\mathrm{1} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{7}}{{R}} \\ $$$${R}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} =\mathrm{58}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}} +\mathrm{2}×\mathrm{58}×\mathrm{92}×\frac{\mathrm{7}}{{R}} \\ $$$${R}^{\mathrm{3}} −\mathrm{11877}{R}−\mathrm{71704}=\mathrm{0} \\ $$$${R}=\mathrm{2}\sqrt{\mathrm{3959}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{35852}}{\mathrm{3959}\sqrt{\mathrm{3959}}}\right) \\ $$$$\:\:\:\:\approx\mathrm{112}.\mathrm{095} \\ $$
Commented by cortano1 last updated on 25/Oct/22
not R = 56 sir ?
$$\mathrm{not}\:\mathrm{R}\:=\:\mathrm{56}\:\mathrm{sir}\:? \\ $$
Commented by mr W last updated on 25/Oct/22
R is shown as diameter.
$${R}\:{is}\:{shown}\:{as}\:{diameter}. \\ $$
Answered by mr W last updated on 24/Oct/22
Method 2  (√((R^2 −7^2 )(R^2 −92^2 )))=7×92+58R  R^4 −(7^2 +92^2 )R^2 +7^2 ×92^2 =7^2 ×92^2 +58^2 R^2 +2×7×92×58R  R^3 −(7^2 +92^2 +58^2 )R−2×7×92×58=0  R^3 −11877R−71704=0  the rest see above.
$$\boldsymbol{{Method}}\:\mathrm{2} \\ $$$$\sqrt{\left({R}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} \right)\left({R}^{\mathrm{2}} −\mathrm{92}^{\mathrm{2}} \right)}=\mathrm{7}×\mathrm{92}+\mathrm{58}{R} \\ $$$${R}^{\mathrm{4}} −\left(\mathrm{7}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}} \right){R}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} ×\mathrm{92}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} ×\mathrm{92}^{\mathrm{2}} +\mathrm{58}^{\mathrm{2}} {R}^{\mathrm{2}} +\mathrm{2}×\mathrm{7}×\mathrm{92}×\mathrm{58}{R} \\ $$$${R}^{\mathrm{3}} −\left(\mathrm{7}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}} +\mathrm{58}^{\mathrm{2}} \right){R}−\mathrm{2}×\mathrm{7}×\mathrm{92}×\mathrm{58}=\mathrm{0} \\ $$$${R}^{\mathrm{3}} −\mathrm{11877}{R}−\mathrm{71704}=\mathrm{0} \\ $$$${the}\:{rest}\:{see}\:{above}. \\ $$

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