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Question Number 192001 by ajfour last updated on 05/May/23

Commented by ajfour last updated on 05/May/23

$F i n d x, i n t e r m s o f a, b, m = \tan θ .$
	Answered by ajfour last updated on 05/May/23
	$say θ=2φ, x=c cos (φ+sin^(−1) ((a−c)/(a+c))+sin^(−1) ((a−b)/(a+b))+φ) =(((a+b)^2 −(b+c)^2 +(c+a)^2 )/(2(a+b)(c+a))) say if φ=0 (((2(√(ac)))/(a+c)))(((2(√(ab)))/(a+b)))−(((a−c)/(a+c)))(((a−b)/(a+b))) =(((a+b)^2 −(b+c)^2 +(c+a)^2 )/(2(a+b)(c+a))) ⇒ 4a(√(bc))−{a^2 −(b+c)a+bc} =a^2 +a(b+c)−bc ⇒ 2a^2 −4a(√(bc))=0 ⇒ if a≠0 then a=2(√(bc)) ✓$
	$s a y θ = 2 ϕ, x = c$ $\cos (ϕ + \sin^{- 1} \frac{a - c}{a + c} + \sin^{- 1} \frac{a - b}{a + b} + ϕ)$ $= \frac{{(a + b)}^{2} - {(b + c)}^{2} + {(c + a)}^{2}}{2 (a + b) (c + a)}$ $s a y i f ϕ = 0$ $(\frac{2 \sqrt{a c}}{a + c}) (\frac{2 \sqrt{a b}}{a + b}) - (\frac{a - c}{a + c}) (\frac{a - b}{a + b})$ $= \frac{{(a + b)}^{2} - {(b + c)}^{2} + {(c + a)}^{2}}{2 (a + b) (c + a)}$ $\Rightarrow 4 a \sqrt{b c} - {a^{2} - (b + c) a + b c}$ $= a^{2} + a (b + c) - b c$ $\Rightarrow 2 a^{2} - 4 a \sqrt{b c} = 0$ $\Rightarrow i f a \neq 0 t h e n$ $a = 2 \sqrt{b c}$ $✓$
	Answered by mr W last updated on 05/May/23

	Commented by mr W last updated on 05/May/23

	$\cos α = \frac{a - b}{a + b}$ $\cos β = \frac{a - x}{a + x}$ $\cos γ = \frac{{(a + b)}^{2} + {(a + x)}^{2} - {(b + x)}^{2}}{2 (a + b) (a + x)} = \frac{a (a + b) + (a - b) x}{a (a + b) + (a + b) x}$ $α + β + γ = 180 + θ$ $γ = 180 + θ - (α + β)$ $l e t δ = α - θ = \cos^{- 1} \frac{a - b}{a + b} - θ$ $\cos γ = - \cos (δ + β) = - \cos δ \cos β + \sin δ \sin β$ $\Rightarrow \frac{a (a + b) + (a - b) x}{a (a + b) + (a + b) x} + (\frac{a - x}{a + x}) \cos δ - \sin δ \sqrt{1 - {(\frac{a - x}{a + x})}^{2}} = 0$
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