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Question-166822

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Question Number 166822 by ajfour last updated on 28/Feb/22
Answered by mr W last updated on 28/Feb/22
Commented by mr W last updated on 28/Feb/22
OA^2 =1^2 +(1+(1/(tan (θ/2))))^2 cos θ=((1/(tan (θ/2)))/(1+(√(1+(1+(1/(tan (θ/2))))^2 )))) ⇒θ≈65.444°
$${OA}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\theta=\frac{\frac{\mathrm{1}}{\mathrm{tan}\:\:\frac{\theta}{\mathrm{2}}}}{\mathrm{1}+\sqrt{\mathrm{1}+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\theta\approx\mathrm{65}.\mathrm{444}° \\ $$
Commented by ajfour last updated on 01/Mar/22
((1−t^2 )/(1+t^2 ))=(1/(t+(√(t^2 +(t+1)^2 ))))=(((√(t^2 +(t+1)^2 ))−t)/((t+1)^2 )) ⇒ {(1−t^2 )(t+1)^2 +t(1+t^2 )}^2 =(1+t^2 )^2 {t^2 +(t+1)^2 } ⇒ (1−t^2 )^2 (t+1)^4 +t^2 (1+t^2 )^2 +2t(1−t^2 )(1+t^2 )(t+1)^2 =t^2 (1+t^2 )^2 +(t+1)^2 (1+t^2 )^2 ⇒ (1−t^2 )^2 (t+1)^2 +2t(1−t^4 )=(1+t^2 )^2 ⇒ t^6 −2t^4 +t^2 +2t−4t^3 +2t^5 +1−2t^2 +t^4 +2t−2t^5 =1+t^4 +2t^2 ⇒ t^5 −2t^3 −4t^2 −3t+4=0 t=tan (θ/2)≈0.6425 ⇒ θ ≈ 65.44°
$$\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{{t}+\sqrt{{t}^{\mathrm{2}} +\left({t}+\mathrm{1}\right)^{\mathrm{2}} }}=\frac{\sqrt{{t}^{\mathrm{2}} +\left({t}+\mathrm{1}\right)^{\mathrm{2}} }−{t}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\left\{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left({t}+\mathrm{1}\right)^{\mathrm{2}} +{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left\{{t}^{\mathrm{2}} +\left({t}+\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$$$ \\ $$$$\Rightarrow\:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left({t}+\mathrm{1}\right)^{\mathrm{4}} +\cancel{{t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:+\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({t}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:=\cancel{{t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }+\left({t}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left({t}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{4}} \right)=\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{4}{t}^{\mathrm{3}} +\cancel{\mathrm{2}{t}^{\mathrm{5}} } \\ $$$$\:\:\:\:+\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} +\cancel{{t}^{\mathrm{4}} }+\mathrm{2}{t}−\cancel{\mathrm{2}{t}^{\mathrm{5}} }=\mathrm{1}+\cancel{{t}^{\mathrm{4}} }+\mathrm{2}{t}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${t}^{\mathrm{5}} −\mathrm{2}{t}^{\mathrm{3}} −\mathrm{4}{t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{4}=\mathrm{0} \\ $$$${t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\approx\mathrm{0}.\mathrm{6425} \\ $$$$\Rightarrow\:\:\theta\:\approx\:\mathrm{65}.\mathrm{44}° \\ $$$$ \\ $$
Commented by Tawa11 last updated on 01/Mar/22
Great sirs
$$\mathrm{Great}\:\mathrm{sirs} \\ $$

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