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# very confusing pythagorean triple problem... stumped me

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The numbers 20, 99, and 101 form a Pythagorean triple. A right triangle has heights x,$$\dfrac{20}{101}$$, and $$\dfrac{99}{101}$$, where x is the shortest height. What is x?

Jun 26, 2019

#1
+4 sin( angle )  =  opposite / hypotenuse

sin( C )  =  $$\frac{20}{101}$$ / $$\frac{101}{101}$$

sin( C  )  =  $$\frac{20}{101}$$

Also...

sin( C )  =  x / $$\frac{99}{101}$$

sin( C )  =  $$\frac{101}{99}x$$

So we can equate both expressions of  sin( C )  and solve for  x

$$\frac{20}{101}\ =\ \frac{101}{99}x\\~\\ \frac{20}{101}\cdot\frac{99}{101}\ =\ x\\~\\ \frac{1980}{10201}\ =\ x$$_

Jun 26, 2019
#2
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Thanks, that turns out to be right Hectictar!

sudsw12  Jun 26, 2019
#3
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Ok, awesome!!! hectictar  Jun 26, 2019
#4
+2   Jun 26, 2019
edited by CPhill  Jun 26, 2019
edited by CPhill  Jun 26, 2019
#5
+3

The numbers $$20$$, $$99$$, and $$101$$ form a Pythagorean triple.
A right triangle has heights $$x$$, $$\dfrac{20}{101}$$, and $$\dfrac{99}{101}$$,

where $$x$$ is the shortest height.
What is $$x$$?

Formula:  $$\begin{array}{|rcll|} \hline x &=& \dfrac{\dfrac{20}{101}\cdot \dfrac{99}{101}}{\dfrac{101}{101}} \\\\ &=& \dfrac{\dfrac{20\cdot 99}{101^2}}{1} \\\\ &=& \dfrac{20\cdot 99}{101^2} \\\\ &=& \dfrac{1980}{10201} \\\\ \mathbf{x} &=& \mathbf{0.19409861778} \\ \hline \end{array}$$ Jun 26, 2019