Triangle identity with two large sides and 1 small side I have been tasked at work to take over a task from a former colleague and I cant wrap my head around the trig proof in his notes
I can prove it works when I use some real numbers, but cant get how to prove it through trig identities/formulas
What I have is the following figure and I know the values of d0 (red line), and alpha (angle).
I am trying to prove that d0 * cos(180 - alpha) == d1 - d2
I do not need to solve for d1 or d2 (unknown), just prove that the difference is equivalent to  d0 * cos(180 - alpha), which I do know the values of.
d1 and d2 are very very large with respect to d0, so alpha1 is numerically almost identical to alpha2
Additionally, if d0 is not parallel with the y-axis we get d0 * cos(180 - alpha + theta) == d1 - d2.
Any help or pointers to write out this proof would be much appreciated.

 A: Based on the assumptions, we can say that $d_1+d_2 \approx 2d_1$.  We apply the law of cosines to the triangle in the top diagram:
$$d_1^2 = d_0^2+d_2^2-2d_0d_1\cos(180-\alpha_1).$$
This rearranges to give
$$d_1^2-d_2^2 = d_0(d_0-2d_1\cos(180-\alpha_1)).$$
Factor the LHS and use $d_1+d_2 \approx 2d_1$ to obtain:
$$2d_1(d_1-d_2) \approx d_0(d_0-2d_1\cos(180-\alpha_1)).$$
Now, because $d_0 << d_1$, $d_0-2d_1\cos(180-\alpha_1) \approx -2d_1\cos(180-\alpha_1)$.
The above becomes
$$2d_1(d_1-d_2) \approx -d_0(2d_1\cos(180-\alpha_1)).$$
Cancel $2d_1$ from both sides.
$$d_1-d_2 \approx -d_0\cos(180-\alpha_1).$$
Notice that this is negative of what you claim.  This makes sense because $180-\alpha_1>90$ so that the right hand side above is (negative)*(negative) = positive, which agrees with the LHS since $d_1>d_2$.  So, the correct result (for $\alpha_1$ acute angle) is:
$$d_1-d_2 \approx -d_0\cos(180-\alpha_1).$$
The second part is the same argument, replacing the angle used in law of cosines with $\alpha_1-\theta$.  Notice that
$$180-(\alpha_1-\theta) = 180-\alpha_1+\theta.$$
Hope this helps!
