enter image description here

\begin{align}\frac z{\sin E}&=\frac y{\sin D}\tag{1}\\ z^2 &= x^2 + y^2 - 2xy\cos E\tag{2}\\ y^2 &= z^2 + x^2 - 2zx\cos D\tag{3}\end{align}

Solve for $z$

My solution is as follow.
enter image description here

I stopped at the equation:

$$(2xy)^2 - (x^2 + y^2 - z^2)^2 = (2zx)^2 - (z^2 + x^2 - y^2)^2$$

Please help me solve for $z$ in the above equation.

Thank you

  • 1
    $\begingroup$ I think, It's impossible. We need more given. $\endgroup$ – Michael Rozenberg May 25 '17 at 7:08
  • $\begingroup$ All data are given. Also, I arrived at the equation: (2xy)^2 - (x^2 + y^2 - z^2)^2 = (2zx)^2 - (z^2 + x^2 - y^2)^2 There is only one unknown in the above equation, which is z. x and y are given, and they are x = 70.64622683 and y = 147.2977643 $\endgroup$ – Chapel Li May 25 '17 at 7:13
  • $\begingroup$ are the angles given? $\endgroup$ – Dr. Sonnhard Graubner May 25 '17 at 7:14
  • $\begingroup$ Angles are not given $\endgroup$ – Chapel Li May 25 '17 at 7:16
  • $\begingroup$ The reason why I stopped at the equation: (2xy)^2 - (x^2 + y^2 - z^2)^2 = (2zx)^2 - (z^2 + x^2 - y^2)^2 is because I don't know how to solve a quartic function. $\endgroup$ – Chapel Li May 25 '17 at 7:21

If you expand your final equation $$(2xy)^2 - (x^2 + y^2 - z^2)^2 = (2zx)^2 - (z^2 + x^2 - y^2)^2$$ you will get an equation which has equal terms in $x^4$, $y^4$, and $z^4$ on both sides. Removing these gets you:

$$2x^2y^2 + 2x^2z^2 + 2y^2z^2 = 2x^2y^2 + 2x^2z^2 + 2y^2z^2$$

which as you can see is always true.

The problem is that all your algebraic manipulations cannot add information, they can just rearrange what you already know. Look at your diagram. You have two sides of a triangle, and nothing else. There are an infinite number of triangles you can make that have two sides of given lengths. You can see this physically - get two sticks, attach their ends, and pivot. Every position gives you a new triangle, with a different third side length.

  • $\begingroup$ Thank you for your excellent explanation. You are a good teacher. $\endgroup$ – Chapel Li May 25 '17 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.