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Now, I got the first one by using the like triangles. This is my work, please tell me if I'm right: $$\frac{t}{h}=\frac{x+t}{r}\Rightarrow x+t=\frac{rt}{h}\Rightarrow x=\frac{rt}{h-t}$$ Now, I figure that $(b)$ must use the same concept (big triangle=little triangle). But I don't know what or how to do so. Please, give me general hints and only start me off, do not solve this problem for me.

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    Please check the last expression. It seems that there is a typo, and an error in calculating.2012-08-10
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    Let me re-work it.2012-08-10
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    When you subtract $t$ from both sides, you need a **common denominator**2012-08-10
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    So, multiply $x+t$ by $\frac{1}{h}$?2012-08-10
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    $x + t = \dfrac{rt}{h} \Rightarrow x = \dfrac{rt}{h} - t = \dfrac{rt}{h} - \dfrac{th}{h} = \cdots$2012-08-10
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    $x=\dfrac{t(r-h)}{h}$ or is it $x=t\left(\dfrac{r-h}{h}\right)$?2012-08-10
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    @AustinBroussard: Doesn't matter, both are right.2012-08-10
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    Austin, both are correct. Sorry... Fell asleep"2012-08-10

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For (b), use similarity to conclude that $$\frac{x}{t}=\frac{r}{w},$$ where $w$ is the remaining side of the little triangle. There remains some work to do.

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    Could I also conclude that: $$w^2=r^2-h^2\Rightarrow w=\sqrt{r^2-h^2}\Rightarrow t=\sqrt{r^2-h^2}$$ $$\dfrac{x}{r}=\dfrac{t}{\sqrt{r^2-h^2}}$$ And then I'd solve that?2012-08-10
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    @GerryMyerson: Thanks, corrected.2012-08-10
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    @AustinBroussard: I had a typo, it is $\frac{x}{t}=\frac{r}{w}$. Your calculation of $w$ is correct, and your equation is correct. Now it is only a small step to the end.2012-08-10
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    Final answer coming to $x=\dfrac{rt}{\sqrt{r^2-h^2}}$?2012-08-10
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    @AustinBroussard: Yes, correct.2012-08-10
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    @AndréNicolas Thanks for the help. Your small comment actually helped a ton! How ironic.2012-08-10