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Question Number 187535 by ajfour last updated on 18/Feb/23

Commented by ajfour last updated on 18/Feb/23

Q.187482

$${Q}.\mathrm{187482} \\ $$

Answered by ajfour last updated on 19/Feb/23

We alraedy have obtained, s=tan α=(R/H) pcos 30°=b ⇒ (p/b)=(2/( (√3))) ((bcos α+p)/(ccos α))=((AE)/(AGcos α))=((2(√(bc)))/(ccot α))+1 ⇒ (b/c)(1+(2/( (√3)cos α)))=(2/(cot α))(√(b/c))+1 say (b/c)=λ tan α=s, cos α=(1/( (√(1+s^2 )))) λ(1+((2(√(1+s^2 )))/( (√3))))−2s(√λ)−1=0 (1/λ)+((2s)/( (√λ)))−(1+((2(√(1+s^2 )))/( (√3))))=0 (1/( (√λ)))=(√(c/b))=−s+(√(s^2 +1+2(√((1+s^2 )/3)))) −−−−−−−−−−−−−−−− examples: for s=(1/( (√2))) ⇒ sin α=(1/( (√3))) (1/( (√λ)))=(√(c/b))=−(1/( (√2)))+(√((3/2)+(√2)))=1 for α=60° (1/( (√λ)))=−(√3)+(√(3+1+(4/( (√3))))) (√(c/b))=−(√3)+2(√(1+(1/( (√3))))) ≈ 0.7798 ⇒ b≈1.6445 for α=30° (√(c/b))=((2(√2)−1)/( (√3))) b≈0.8974

$${We}\:{alraedy}\:\:{have}\:{obtained}, \\ $$$${s}=\mathrm{tan}\:\alpha=\frac{{R}}{{H}} \\ $$$${p}\mathrm{cos}\:\mathrm{30}°={b}\:\:\:\Rightarrow\:\:\frac{{p}}{{b}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{b}\mathrm{cos}\:\alpha+{p}}{{c}\mathrm{cos}\:\alpha}=\frac{{AE}}{{AG}\mathrm{cos}\:\alpha}=\frac{\mathrm{2}\sqrt{{bc}}}{{c}\mathrm{cot}\:\alpha}+\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{b}}{{c}}\left(\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha}\right)=\frac{\mathrm{2}}{\mathrm{cot}\:\alpha}\sqrt{\frac{{b}}{{c}}}+\mathrm{1} \\ $$$${say}\:\frac{{b}}{{c}}=\lambda\:\:\:\mathrm{tan}\:\alpha={s},\:\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }} \\ $$$$\lambda\left(\mathrm{1}+\frac{\mathrm{2}\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{2}{s}\sqrt{\lambda}−\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\lambda}+\frac{\mathrm{2}{s}}{\:\sqrt{\lambda}}−\left(\mathrm{1}+\frac{\mathrm{2}\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\lambda}}=\sqrt{\frac{{c}}{{b}}}=−{s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}\sqrt{\frac{\mathrm{1}+{s}^{\mathrm{2}} }{\mathrm{3}}}} \\ $$$$−−−−−−−−−−−−−−−− \\ $$$${examples}: \\ $$$${for}\:\:\:{s}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\:\Rightarrow\:\:\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\lambda}}=\sqrt{\frac{{c}}{{b}}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\mathrm{2}}}=\mathrm{1} \\ $$$${for}\:\alpha=\mathrm{60}° \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\lambda}}=−\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}+\mathrm{1}+\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}} \\ $$$$\:\:\:\:\sqrt{\frac{{c}}{{b}}}=−\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\:\mathrm{0}.\mathrm{7798} \\ $$$$\Rightarrow\:\:\:{b}\approx\mathrm{1}.\mathrm{6445} \\ $$$${for}\:\:\alpha=\mathrm{30}° \\ $$$$\:\:\:\:\:\:\sqrt{\frac{{c}}{{b}}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:{b}\approx\mathrm{0}.\mathrm{8974} \\ $$

Commented by ajfour last updated on 19/Feb/23

Thanks Sir, edited.

$${Thanks}\:{Sir},\:{edited}. \\ $$

Commented by mr W last updated on 19/Feb/23

perfect solution sir!

$${perfect}\:{solution}\:{sir}! \\ $$

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