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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. –  The Chaz 2.0 Aug 10 '12 at 5:05
    
Let me re-work it. –  Austin Broussard Aug 10 '12 at 5:06
    
When you subtract $t$ from both sides, you need a common denominator –  The Chaz 2.0 Aug 10 '12 at 5:09
    
So, multiply $x+t$ by $\frac{1}{h}$? –  Austin Broussard Aug 10 '12 at 5:10
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@AustinBroussard: Doesn't matter, both are right. –  André Nicolas Aug 10 '12 at 5:49
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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? –  Austin Broussard Aug 10 '12 at 5:44
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@GerryMyerson: Thanks, corrected. –  André Nicolas Aug 10 '12 at 5:44
    
@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. –  André Nicolas Aug 10 '12 at 5:47
    
Final answer coming to $x=\dfrac{rt}{\sqrt{r^2-h^2}}$? –  Austin Broussard Aug 10 '12 at 5:48
    
@AustinBroussard: Yes, correct. –  André Nicolas Aug 10 '12 at 5:51
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