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I have two right triangles that share the same hypotenuse. Can the length of Xb be found using just the other lengths shown (X, L, Y)? I have only been able to find it by using a combination of the asine, atangent, and tangent of the resulting angles, which seems like a much too complicated and non-elegant solution. Thanks for your help!

triangles

user154739
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1 Answers1

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Yes. It can be done but as you mentioned it is non-elegant.

enter image description here

Perform step (1) to (4).

Then by power of a point, do $(3).(4) = (1).X_b$

After simplification, you should get the following quadratic equation in X(b) , if my calculation did not make any mistake.

$X_{b}^2 (L^2 – Y^2) + 2Y^2XX_{b} –(X^2 + Y^2 – L^2)Y^2 = 0$

Edit: Let $K = X^2 + Y^2 – L^2$.

Then, $[√K – √(X_b^2 + Y^2)][√(X_b^2 + Y^2)] = [X.X_b – X_b^2]$

$√K√(X_b^2 + Y^2) – (X_b^2 + Y^2) = [X.X_b – X_b^2]$

$√K√(X_b^2 + Y^2) = [X.X_b – X_b^2] + (X_b^2 + Y^2)$

$√K√(X_b^2 + Y^2) = [X.X_b + Y^2]$

$K(X_b^2 + Y^2) = X^2.X_b^2 + 2XY^2X_b+ Y^4$

$(X^2 – K)X_b^2+ 2XY^2X_b + Y^2(Y^2 – K)= 0$

Putting K back to get

$(L^2 – Y^2)X_b^2 + 2XY^2X_b + Y^2(L^2 – X^2)= 0$

Mick
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  • Thanks so much Mick for the quick and knowledgeable reply! It will take me a little while to confirm your derivation (I can't seem to eliminate one of the square root terms through squaring/rearranging etc..). As a test I tried to verify the formula by comparing it to the Xb dimension that I measure in CAD, but I get a different answer. Thanks again. Dave. – user154739 Jun 02 '14 at 18:28
  • @user154739 You are welcome. I did make a small mistake in my calculation. I have the detail workout added to to my post in the edited section. After all, the resultant equation is not that ugly. – Mick Jun 03 '14 at 03:05
  • Great! Thanks for the update and for adding the derivation! – user154739 Jun 03 '14 at 12:36