What trigonometric identity makes the method of triangulation work?

Question

I've read the article on Wikipedia, but I don't get how to construct the relationships between sides and angles to reach a solution for the distance between two points.

All the other sites I read either just tell you to scale the triangle in a piece of paper or they use a right-angled triangle (making one of the angles you measure be 90 degrees).

I'd really appreciate any help. Thanks!

Answer 1

In the figure, $x = AB, y = AD, z = AC, h = CD$, where $\overline{CD}$ is perpendicular to $\overline{AB}$.

A common application of triangulation: You want to measure the distance from $A$, where you are to a distance object $C$. So you move over a known distance $x$ (usually much smaller than the distance to $C$) to another point $B$ where you can see both $A$ and $C$, then measure the angles at $A$ and $B$ (which I will denote by the same letters).

So you know the angles $A$ and $B$, and the distance $x$. And you want to find $z$. By the Angle-Side-Angle theorem, $A,B, x$ are enough to completely determine the triangle. Drop the height from $C$ onto $\overline{AB}$ intersecting in $D$. $y$ is the distance from $A$ to $D$, which means that the distance from $B$ to $D$ is $(x - y)$. Now we have two right triangles. Since the tangent is opposite over adjacent, we have two expressions for $h$:

$$h = y\tan A \qquad\text{and}\qquad h = (x - y)\tan B$$

Equating these and solving for $y$ gives: $$y = \frac {x\tan B}{\tan A + \tan B}$$ The left triangle also gives $y = z\cos A$, or $z = \frac y{\cos A}$. Therefore $$z = \frac {x\tan B}{\cos A (\tan A + \tan B)} = \frac{x\sin B}{\sin(A + B)}$$

So from the three measurements you made locally, you are able to compute the distance to the far off point $C$.

Answer 2

You ask, concerning the Wikipedia article, "Why do we conclude $\ell =d/\tan(\alpha) + d/\tan(\beta)$?"

Rather than draw yet another figure, let me just note that $d$ in the Wikipedia article is the same as $h$ in the figure in Paul Sinclair's answer; $\alpha$ is the angle at $A$, and $\beta$ is the angle at $B$.

In Paul's figure, the point $D$ divides the segment $\overline {AB}$ (which has length $x$) into two pieces of lengths $y$ and $x - y$ respectively. This gives us (quoting from that answer), \begin{align} h &= y\tan A, \\ h &= (x - y)\tan B. \end{align}

Equivalently, \begin{align} y &= \frac{h}{\tan A}, \tag 1\\ x - y &= \frac{h}{\tan B}. \tag 2 \end{align} Therefore $$ x = y + (x - y) = \frac{h}{\tan A} + \frac{h}{\tan B}, \tag 3 $$ that is, segment $\overline {AB}$ consists of two pieces that are legs of right triangles with angles $A$ and $B$, respectively, and the lengths of those legs can be determined by trigonometry. This is the same as the equation you asked about except for the labeling of the figure: compared to the figure in the Wikipedia article, Equation $(3)$ uses $x$ for the length $\ell$ and $A$ and $B$ for the angles $\alpha$ and $\beta$.

Wikipedia did not explicitly state Equations $(1)$ and $(2)$. Apparently we are supposed to infer them from the figure and write the results directly into Equation $(3)$.

Once you have the length of $\overline {CD}$, you can use that, the known angles, and the sine function to determine the lengths of $\overline {AC}$ and $\overline {BC}$. The Wikipedia article seems to assume you can set up the formulas; it does not give them explicitly.