Question-202684

Question Number 202684 by ajfour last updated on 31/Dec/23

Answered by mr W last updated on 02/Jan/24

cos α=((s^2 +q^2 −((s/2))^2 )/(2sq))=((3s^2 +4q^2 )/(8sq)) cos β=(p/(2s))=((s−q)/(2s))=sin α (((s−q)/(2s)))^2 +(((3s^2 +4q^2 )/(8sq)))^2 =1 32q^4 −32sq^3 −24s^2 q^2 +9s^4 =0 let λ=(q/s) λ^4 −λ^3 −((3λ^2 )/4)+(9/(32))=0 ⇒λ≈0.5306, 1.4285 (>1, rejected) (q/p)=(λ/(1−λ))≈((0.5306)/(1−0.5306))≈1.1304 ✓

$$\mathrm{cos}\:\alpha=\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}{sq}}=\frac{\mathrm{3}{s}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{2}} }{\mathrm{8}{sq}} \\ $$$$\mathrm{cos}\:\beta=\frac{{p}}{\mathrm{2}{s}}=\frac{{s}−{q}}{\mathrm{2}{s}}=\mathrm{sin}\:\alpha \\ $$$$\left(\frac{{s}−{q}}{\mathrm{2}{s}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}{s}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{2}} }{\mathrm{8}{sq}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{32}{q}^{\mathrm{4}} −\mathrm{32}{sq}^{\mathrm{3}} −\mathrm{24}{s}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{9}{s}^{\mathrm{4}} =\mathrm{0} \\ $$$${let}\:\lambda=\frac{{q}}{{s}} \\ $$$$\lambda^{\mathrm{4}} −\lambda^{\mathrm{3}} −\frac{\mathrm{3}\lambda^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{9}}{\mathrm{32}}=\mathrm{0} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{5306},\:\mathrm{1}.\mathrm{4285}\:\left(>\mathrm{1},\:{rejected}\right) \\ $$$$\frac{{q}}{{p}}=\frac{\lambda}{\mathrm{1}−\lambda}\approx\frac{\mathrm{0}.\mathrm{5306}}{\mathrm{1}−\mathrm{0}.\mathrm{5306}}\approx\mathrm{1}.\mathrm{1304}\:\checkmark \\ $$

Commented by mr W last updated on 01/Jan/24