I have the following setup:(We know every point, length and angle illustrated in the figure, expect for $z$)
I am then trying to figure out the length $z$ going from the point $X$ to the middle of the baseline.
Here is what I have managed to think myself:
The angle at $C$ and $C'$ in the small triangles are the same as the angle at $X$ in the triangle formed by $\overrightarrow{CXF}$ and $\overrightarrow{C'XF}$ where $F$ is a point in the middle of the baseline. That is, $\theta_1 = \theta_2$. I then thought that applying the definitions of sine, cosine or tangent might work but I would only know the opposite line of the triangle so wasn't a success. I got stuck here :(
Can anybody help me out?
As a matter of fact, I know that the answer should be $z = - \frac{f B}{x' - x}$ but do not know how it is derived, which is what I am interested in.
$\endgroup$ 41 Answer
$\begingroup$Drop a perpendicular from $X$ to the baseline. Call the base of the altitude $Y$.
We form two similar triangles, hence we have the equal ratios:
$$\frac fx = \frac z{CY}, \quad \frac f{x'} = \frac z{C'Y}$$
We also have $B = CY + C'Y = \dfrac {xz}f + \dfrac {x'z}f = z\left(\dfrac{x+x'}f\right)$.
Hence $z = \dfrac {fB}{x+ x'}$.
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