How do I find a third side of a triangle with two sides and a bisecting line segment?

Question

I am using a laser range finder to calculate the height of a second story wall. I have a fixed point and three separate lengths hitting the top, the bottom, and an indeterminate point on the wall. With the exact lengths of all three line segments, how do I find the length of the wall? Note: I am unable to create a right triangle due to the height and the uneven wall underneath.

Answer 1

Let $a = AX, b = XB$, where $A, B$ are the lower end and upper ends of the segment whose length is what you are solving for. Using the bisector property of triangle, and the cosine formula you have: $\dfrac{96}{109} = \dfrac{a}{b}, \dfrac{103^2+96^2-a^2}{2\cdot 103\cdot 96}=\dfrac{103^2+109^2-b^2}{2\cdot 103\cdot 109}$. Can you solve this system and your answer is $a+b$.

Answer 2

If the middle line intersects the wall at an arbitrary point, and not the midpoint, you can't solve for X. For an example I'll use the "Law of Cosines", which is

$c^2 = a^2 + b^2 -2\cdot a\cdot b\cdot cos(\theta)$

If we let the 2 angles be $15^{\circ}$and $10^{\circ}$, then X=46.3056.

If we let the 2 angles be $20^{\circ}$and $15^{\circ}$, then X=64.1705.

Answer 3

Results from DeepSea's calculation:

Add the two segment lengths and simplify

$$ X=(1+\frac{r}{p}) \sqrt{\dfrac{p^3-pq^2+p^2r+q^2r-pr^2}{r}} $$

$$X =(1+\frac{r}{p}) \sqrt{\dfrac{p}{r}(p^2-q^2)+(p^2+q^2-pr)} $$

The wall need not be vertical, the scheme is even valid for three collinear points on a ceiling.