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Question Number 210694 by ajfour last updated on 16/Aug/24

	Answered by mr W last updated on 17/Aug/24

	Commented by mr W last updated on 17/Aug/24

	$R = \sqrt{a^{2} + b^{2}}$ $\frac{\sin α}{b} = \frac{\sin (\frac{π}{2} - θ + α)}{R} = \frac{\cos θ \cos α + \sin θ \sin α}{R}$ $\frac{R}{b} = \frac{\cos θ}{\tan α} + \sin θ$ $\Rightarrow \frac{1}{\tan α} = (\frac{R}{b} - \sin θ) \frac{1}{\cos θ}$ $\frac{\sin α}{a} = \frac{\sin (θ + α)}{R} = \frac{\sin θ \cos α + \cos θ \sin α}{R}$ $\frac{R}{a} = \frac{\sin θ}{\tan α} + \cos θ$ $\Rightarrow \frac{1}{\tan α} = (\frac{R}{a} - \cos θ) \frac{1}{\sin θ}$ $(\frac{R}{b} - \sin θ) \frac{1}{\cos θ} = (\frac{R}{a} - \cos θ) \frac{1}{\sin θ}$ $\frac{R}{a} \cos θ - \frac{R}{b} \sin θ = \cos 2 θ$ $\Rightarrow \frac{\cos θ}{\cos ϕ} - \frac{\sin θ}{\sin ϕ} = \cos 2 θ$ $L_{m i n} = \sqrt{a^{2} + R^{2} - 2 a R \cos θ} + \sqrt{b^{2} + R^{2} - 2 b R \sin θ}$ $\frac{L_{m i n}}{R} = \sqrt{1 + \cos^{2} ϕ - 2 \cos ϕ \cos θ} + \sqrt{1 + \sin^{2} ϕ - 2 \sin ϕ \sin θ}$
	Commented by mr W last updated on 17/Aug/24

	$a n e x a c t s o l u t i o n s e e m s n o t t o b e$ $p o s s i b l e .$
	Commented by ajfour last updated on 17/Aug/24

	$\frac{\sin α}{b} = \frac{\sin (\frac{π}{2} + θ - α)}{R} = \frac{\cos θ \cos α - \sin θ \sin α}{R}$ $\frac{R}{a} = \frac{\sin θ}{\tan α} - \cos θ$ ${(\frac{\sin θ}{m} - \frac{R}{a})}^{2} = 1 - \sin^{2} θ$ $L_{m i n} = \sqrt{a^{2} + R^{2} - 2 a R \cos θ} + \sqrt{b^{2} + R^{2} - 2 b R \sin θ}$ $= \sqrt{a^{2} + R^{2} - 2 R (\frac{a \sin θ}{m} - R)}$ $+ \sqrt{b^{2} + R^{2} - 2 b R \sin θ}$ $= \sqrt{a^{2} + 3 R^{2} - \frac{2 a R}{m} \sin θ} + \sqrt{b^{2} + R^{2} - 2 b R \sin θ}$ $. . . . . . .$
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