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Question Number 186830 by ajfour last updated on 11/Feb/23

Commented by ajfour last updated on 11/Feb/23

$F i n d x .$
	Answered by ajfour last updated on 11/Feb/23
	$((sin (φ−θ))/x^2 )=((sin φ)/x^3 )=((sin φ)/(x+c)) ⇒ x=((sin φ)/(sin (φ−θ))) Also ((sin φ)/(x+c))==((sin (φ−θ))/x^2 ) ⇒ ((sin φ)/(((sin φ)/(sin (φ−θ)))+c))=((sin^3 (φ−θ))/(sin^2 φ)) ⇒ sin^3 φ=sin^2 (φ−θ){sin φ+csin (φ−θ)} .....$
	$\frac{\sin (ϕ - θ)}{x^{2}} = \frac{\sin ϕ}{x^{3}} = \frac{\sin ϕ}{x + c}$ $\Rightarrow x = \frac{\sin ϕ}{\sin (ϕ - θ)}$ $A l s o \frac{\sin ϕ}{x + c} == \frac{\sin (ϕ - θ)}{x^{2}}$ $\Rightarrow \frac{\sin ϕ}{\frac{\sin ϕ}{\sin (ϕ - θ)} + c} = \frac{\sin^{3} (ϕ - θ)}{\sin^{2} ϕ}$ $\Rightarrow$ $\sin^{3} ϕ = \sin^{2} (ϕ - θ) {\sin ϕ + c \sin (ϕ - θ)}$ $. . . . .$
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