Find one side of a right triangle when you know part of the other side and two angles?

Question

Let's say a boat is anchored on the sea, with the angle $a$ between the surface and the anchor line. The boat then moves a distance $X$ towards the anchor point, whereby the angle changes to $b$. How can I from these known parameters calculate the depth of the anchor point?

I did some juggling with $\tan$ and came up with

$$Y = \frac X{\tan(90-a) - \tan(90-b)}$$

Is this correct?

I did some testing and it seemed to work out, but I'm not entirely sure that I'm testing it correctly. Was hoping somebody could confirm the formula or suggest a better one.

(I forgot the right angle marker in the top right corner of the figure!)

Answer 1

Let the dotted horizontal line have length $T$. Then we have $$\tan b=\frac YT\implies T=\frac Y{\tan b}$$ and $$\tan a=\frac Y{X-T}\implies X\tan a-\frac Y{\tan b}\tan a=Y$$ so $$Y=\frac{X\tan a}{1+\frac{\tan a}{\tan b}}=\frac{X\tan a\tan b}{\tan a+\tan b}$$

Answer 2

Well, my jiggery pokery gives me that

$\tan a = \frac {Y}{X + k}$ and $\tan b = \frac {Y}{k}$

$k$ is an unknown value. $k = \frac Y{\tan b} = \cot b Y$ so

$\tan a = \frac Y{X + \cot b Y}$

$X\tan a + \tan a\cot b Y = Y$

$Y(1-\tan a\cot b) = X\tan a$

$Y = \frac {X\tan a}{1- \tan a\cot b}$

$Y = \frac X{\cot a - \cot b}$

$= \frac X{\tan(90 - a) - \tan(90 -b)}$.

So seems fine to me.

....

You and I both probably been a little more direct.

$M = X + k = Y\times \cot a$

$k = Y\times \cot b$

So $X = M - k = Y(\cot a - \cot b)$ and

$Y = \frac X{\cot a - \cot b}$.