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Question Number 90058 by mr W last updated on 21/Apr/20

Commented by mr W last updated on 21/Apr/20

$f i n d \frac{b}{a} = ? i f Δ P Q R i s e q u i l a t e r a l$
	Answered by ajfour last updated on 21/Apr/20
	$Q(acos θ,bsin θ) ((bsin θ)/(a−acos θ))=(√3) tan (θ/2)=(b/(a(√3))) eq. of BQ: y=((b(1+sin θ))/(acos θ))x−b (b/a)(((1+sin θ))/(cos θ))=(√3) (({1+((2((b/(a(√3)))))/(1+(b^2 /(3a^2 ))))})/((((1−(b^2 /(3a^2 )))/(1+(b^2 /(3a^2 )))))))=((a(√3))/b) let (b/a)=x 1+(x^2 /3)+(2/(√3))x=((√3)/x)(1−(x^2 /3)) x^3 +3(√3)x^2 +3x−3(√3) = 0 x=(b/a)≈ 0.71744$
	$Q (a \cos θ, b \sin θ)$ $\frac{b \sin θ}{a - a \cos θ} = \sqrt{3}$ $\tan \frac{θ}{2} = \frac{b}{a \sqrt{3}}$ $e q . o f B Q : y = \frac{b (1 + \sin θ)}{a \cos θ} x - b$ $\frac{b}{a} \frac{(1 + \sin θ)}{\cos θ} = \sqrt{3}$ $\frac{{1 + \frac{2 (\frac{b}{a \sqrt{3}})}{1 + \frac{b^{2}}{3 a^{2}}}}}{(\frac{1 - \frac{b^{2}}{3 a^{2}}}{1 + \frac{b^{2}}{3 a^{2}}})} = \frac{a \sqrt{3}}{b}$ $l e t \frac{b}{a} = x$ $1 + \frac{x^{2}}{3} + \frac{2}{\sqrt{3}} x = \frac{\sqrt{3}}{x} (1 - \frac{x^{2}}{3})$ $x^{3} + 3 \sqrt{3} x^{2} + 3 x - 3 \sqrt{3} = 0$ $x = \frac{b}{a} \approx 0.71744$
	Commented by mr W last updated on 21/Apr/20

	$t h a n k s s i r!$
	Commented by peter frank last updated on 21/Apr/20

	$h e l p Q n 90030$
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