If you have a square inscribed inside of a right angled triangle. Call the sides of the square $k$. The hypotenuse is $z$. How would you express either of the two other sides of the triangle in terms of $z$ and $k$? The corner of the square touches the hypotenuse-there isn't a side of the square against the hypotenuse.
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Does another corner of the square coincide with the right angle of the triangle? – Henry May 30 '20 at 20:15
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Hint:
Call one angle of the right triangle $\theta$. The square produces two more smaller right triangles whose hypotenuses add up to $z$, i.e. $$k\sec\theta+k\csc \theta =z \\ \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\frac zk \\ \frac{1+2t}{t^2}=\frac{z^2}{k^2}$$ where $t=\sin\theta \cos\theta$. You can solve for $t$ and consequently $\theta$ using this equation. Once you have $\theta$, the required sides will be $k+k\tan\theta$ and $k+k\cot\theta$.
Vishu
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I'm not sure how to solve for t and theta. Could you provide this solution? – tomm0334 May 30 '20 at 20:37
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@tomm0334 Can you solve the quadratic equation for $t$? You have to remember to take the positive root. After you have obtained $t$, $$\sin\theta \cos\theta =t \implies \frac 12 \cdot 2\sin\theta\cos\theta=t \implies \sin 2\theta =2t \implies \theta =\frac 12 \sin^{-1}(2t)$$ – Vishu May 30 '20 at 20:45
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I get a math error. I got t=4. When I put it in, I get math error on the calculator. – tomm0334 May 30 '20 at 21:06
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@tomm0334 $t=4$ is impossible. Why don’t you try calculating it by hand? – Vishu May 31 '20 at 09:47
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k + k tan(0.5 sin^(-1)((4 k^2 + 2 sqrt(4 k^4 + 4 z^2 k^2))/(2 z^2))) I got that as my final answer-is this correct? I got t as 2k^2 + sqrt(4 k^4 + 4 z^2 k^2))/(2 z^2) – tomm0334 May 31 '20 at 20:17
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Hint:
If the other sides are $x$ and $y$ then you have $$\sqrt{x^2+y^2}=z$$ $$\sqrt{(x-k)^2+k^2} +\sqrt{(y-k)^2+k^2} =z$$
which are two equations in the two unknowns. You should expect multiple solutions and so should check for spurious solutions introduced by squaring
Henry
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