# Help with finding the length of one of the sides of a triangle \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. 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

• I think, It's impossible. We need more given. – Michael Rozenberg May 25 '17 at 7:08
• 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 – Chapel Li May 25 '17 at 7:13
• are the angles given? – Dr. Sonnhard Graubner May 25 '17 at 7:14
• Angles are not given – Chapel Li May 25 '17 at 7:16
• 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. – 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:
$$2x^2y^2 + 2x^2z^2 + 2y^2z^2 = 2x^2y^2 + 2x^2z^2 + 2y^2z^2$$