I have the problem described in this image (not to scale): http://i.imgur.com/owWVUmj.png
A (partial cathetus), B (small, vertical cathetus) and C (which forms a 90ยบ angle with hypotenuse) are known lengths (A=3, B=8, C=2 meters, inches, or whatever). I need to obtain the exact values of all sides and angles. The most important length I need is X, but I could calculate it myself once I know the at least one of the angles or lengths.
Could someone please tell me how to calculate at least one of the sides or angles?
Thank you very much.
Best Answer
Let $Y$ be the rest of the leg on which $A$ lies. Then $$\begin{align} \frac{C^2}{Y^2} = \frac{B^2}{B^2 + (A+Y)^2} &\iff B^2Y^2 = C^2(B^2 + A^2 + 2AY + Y^2)\\ &\iff (B^2 - C^2)Y^2 - 2AC^2Y - C^2(A^2 + B^2) = 0.\end{align}$$
Using the quadratic formula,
$$\begin{align} Y & = \dfrac{2AC^2 \pm \sqrt{4A^2C^4 + 4(B^2-C^2)C^2(A^2+B^2)}}{2(B^2-C^2)} \\ & = \dfrac{AC^2 \pm \sqrt{A^2C^4 + A^2B^2C^2 + B^4C^2 - A^2C^4 - B^2C^4}}{B^2-C^2} \\ &= \dfrac{AC^2 \pm C \sqrt{A^2B^2 + B^4 - B^2C^2}}{B^2-C^2} \\ &= \dfrac{AC^2 \pm BC \sqrt{A^2 + B^2 - C^2}}{B^2-C^2}. \end{align}$$ We know $B>C$ since $C$'s triangle is similar to the overall triangle and embedded in it, so the radical is always real. Also $BC \sqrt{A^2 + B^2 - C^2} > ABC > AC^2$, and $Y$ is positive, so the plus case gives us the only solution:
$$Y = \dfrac{AC^2 + BC \sqrt{A^2 + B^2 - C^2}}{B^2-C^2}.$$
Then $\dfrac{Y}{A+Y} = \dfrac{X}{B}$, so $X = \dfrac{BY}{A+Y}$.