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.
Express $x$ in terms of the other variables.
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algebra-precalculus
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0Please check the last expression. It seems that there is a typo, and an error in calculating. – 2012-08-10
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0Let me re-work it. – 2012-08-10
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0When you subtract $t$ from both sides, you need a **common denominator** – 2012-08-10
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0So, multiply $x+t$ by $\frac{1}{h}$? – 2012-08-10
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0$x + t = \dfrac{rt}{h} \Rightarrow x = \dfrac{rt}{h} - t = \dfrac{rt}{h} - \dfrac{th}{h} = \cdots$ – 2012-08-10
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0$x=\dfrac{t(r-h)}{h}$ or is it $x=t\left(\dfrac{r-h}{h}\right)$? – 2012-08-10
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1@AustinBroussard: Doesn't matter, both are right. – 2012-08-10
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0Austin, both are correct. Sorry... Fell asleep" – 2012-08-10
1 Answers
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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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0Could 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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1@GerryMyerson: Thanks, corrected. – 2012-08-10
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0@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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0Final answer coming to $x=\dfrac{rt}{\sqrt{r^2-h^2}}$? – 2012-08-10
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0@AustinBroussard: Yes, correct. – 2012-08-10
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0@AndréNicolas Thanks for the help. Your small comment actually helped a ton! How ironic. – 2012-08-10