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Question Number 210524 by hardmath last updated on 11/Aug/24

Commented by hardmath last updated on 11/Aug/24

$$\mathbf{r}=?$$

Answered by mr W last updated on 12/Aug/24

Commented by mr W last updated on 12/Aug/24

$$\alpha ={\sin }^{-1}\frac{r}{a}$$$$\beta ={\sin }^{-1}\frac{r}{b}$$$$\gamma ={\sin }^{-1}\frac{r}{c}$$$$\delta ={\sin }^{-1}\frac{r}{d}$$$$2\alpha +2\gamma =\pi \Rightarrow \alpha +\gamma =\frac{\pi }{2}\Rightarrow \sin \alpha =\cos \gamma$$$$\Rightarrow \frac{r}{a}=\sqrt{1-{\left(\frac{r}{c}\right)}^2}$$$$\Rightarrow \frac{r^2}{a^2}=1-\frac{r^2}{c^2}$$$$\Rightarrow r=\frac{ac}{\sqrt{a^2+c^2}}$$$$or$$$$\Rightarrow r=\frac{bd}{\sqrt{b^2+d^2}}$$

Commented by hardmath last updated on 12/Aug/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}\mathrm{cool}$$

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