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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

	${O A}^{2} = 1^{2} + {(1 + \frac{1}{\tan \frac{θ}{2}})}^{2}$ $\cos θ = \frac{\frac{1}{\tan \frac{θ}{2}}}{1 + \sqrt{1 + {(1 + \frac{1}{\tan \frac{θ}{2}})}^{2}}}$ $\Rightarrow θ \approx 65.444 °$
	Commented by ajfour last updated on 01/Mar/22
	$((1−t^2 )/(1+t^2 ))=(1/(t+(√(t^2 +(t+1)^2 ))))=(((√(t^2 +(t+1)^2 ))−t)/((t+1)^2 )) ⇒ {(1−t^2 )(t+1)^2 +t(1+t^2 )}^2 =(1+t^2 )^2 {t^2 +(t+1)^2 } ⇒ (1−t^2 )^2 (t+1)^4 +t^2 (1+t^2 )^2 +2t(1−t^2 )(1+t^2 )(t+1)^2 =t^2 (1+t^2 )^2 +(t+1)^2 (1+t^2 )^2 ⇒ (1−t^2 )^2 (t+1)^2 +2t(1−t^4 )=(1+t^2 )^2 ⇒ t^6 −2t^4 +t^2 +2t−4t^3 +2t^5 +1−2t^2 +t^4 +2t−2t^5 =1+t^4 +2t^2 ⇒ t^5 −2t^3 −4t^2 −3t+4=0 t=tan (θ/2)≈0.6425 ⇒ θ ≈ 65.44°$
	$\frac{1 - t^{2}}{1 + t^{2}} = \frac{1}{t + \sqrt{t^{2} + {(t + 1)}^{2}}} = \frac{\sqrt{t^{2} + {(t + 1)}^{2}} - t}{{(t + 1)}^{2}}$ $\Rightarrow {(1 - t^{2}) {(t + 1)}^{2} + t (1 + t^{2})}^{2}$ $= {(1 + t^{2})}^{2} {t^{2} + {(t + 1)}^{2}}$ $\Rightarrow {(1 - t^{2})}^{2} {(t + 1)}^{4} + t^{2} {(1 + t^{2})}^{2}$ $+ 2 t (1 - t^{2}) (1 + t^{2}) {(t + 1)}^{2}$ $= t^{2} {(1 + t^{2})}^{2} + {(t + 1)}^{2} {(1 + t^{2})}^{2}$ $\Rightarrow$ ${(1 - t^{2})}^{2} {(t + 1)}^{2} + 2 t (1 - t^{4}) = {(1 + t^{2})}^{2}$ $\Rightarrow t^{6} - 2 t^{4} + t^{2} + 2 t - 4 t^{3} + 2 t^{5}$ $+ 1 - 2 t^{2} + t^{4} + 2 t - 2 t^{5} = 1 + t^{4} + 2 t^{2}$ $\Rightarrow$ $t^{5} - 2 t^{3} - 4 t^{2} - 3 t + 4 = 0$ $t = \tan \frac{θ}{2} \approx 0.6425$ $\Rightarrow θ \approx 65.44 °$
	Commented by Tawa11 last updated on 01/Mar/22

	$Great sirs$
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