# What is the Maximum Theoretical Angle a Grand Piano Could be Held At?

Out of curiosity, I wondered why grand pianos have their stand at the length and position that they are made at. I never could find an answer so I decided to try to solve for the maximum angle (B) the lid could be opened to given the length of the prop (b) and the distance out to the prop (a).

I used law of cosines to determine the value of B in terms of a, b, and angle C. I didn't use c because that value is a dependent variable so instead I rewrote c in terms of C, a, and b. I got that $$B = \cos^{-1}(\frac{a-b\cos(C)}{\sqrt{a^2+b^2-2ab\cos(C)}})$$. This then let me solve for the maximum C by taking the derivative and solving for $$0$$. I got $$C = \tan^{-1}(\frac a b)$$.

I initially thought that angle A would always be perpendicular to c but this doesn't seem to be the case since after substituting C into the equation for B, you can determine that (in radians) $$A = \pi-B-C$$, so $$A = \pi-\tan^{-1}(\frac a b)-\cos^{-1}(\frac{x-y\cos(\tan^{-1}(\frac a b))}{\sqrt{a^2+b^2-2ab\cos(\tan^{-1}(\frac a b))}})$$.

This clearly does not equal $$\frac \pi 2$$ for every $$a$$ and $$b$$. Could someone double check my math and see if this result is accurate?

• According to Wolfram Alpha, $C=\arccos(b/a)$, which implies that $A$ is a right angle. I think there must be an error in your calculations, but without having seen the details, we can't say what the error is. Commented Jul 17 at 18:13
• How did you find this result? What did you ask Wolfram Alpha to find? If you assume that A is a right angle, C would be equal to $\cos^{-1}(b/a)$ but wouldn't that result be different if A is not a right angle? That is why I had the equation B(C,a,b) so I could solve for C and find what A was. I don't know if this makes any sense but that was my train of thought trying to figure this out. Commented Jul 17 at 18:21
• I asked first for the derivative of your expression for $B$, then I chose the simplest answer it gave and asked it to solve for that quantity equal to zero. Basically the same steps you would presumably have done, except that I let the computer do all the hard work. Commented Jul 17 at 18:28

One strategy for solving problems like this is to solve a simpler form of the problem if you can. Note that

$$\tan B = \frac{b\sin C}{a - b\cos C},$$

which is much simpler than the expression for $$\cos B$$. Also note that in order to maximize $$B$$ it is sufficient to maximize $$\tan B$$. So now we have the much simpler problem,

$$\frac{\mathrm d}{\mathrm dC} \frac{b\sin C}{a - b\cos C} = 0.$$

We find that

$$\frac{\mathrm d}{\mathrm dC} \frac{b\sin C}{a - b\cos C} = \frac{b (a \cos C - b)}{(a - b \cos C)^2},$$

so we want $$a \cos C - b = 0$$, hence $$C = \arccos\frac ba$$. From this it follows that the triangle is a right triangle with right angle $$A$$.

Just for the sake of completeness:

For the method in the question, Wolfram Alpha says that

$$\frac{\mathrm d}{\mathrm dC} \cos^{-1}\left(\frac{a-b\cos(C)}{\sqrt{a^2+b^2-2ab\cos(C)}}\right) = \frac{\lvert b\rvert \sin(C) (a \cos(C) - b)} {\lvert\sin(C)\rvert (a^2 - 2 a b \cos(C) + b^2)}$$ for real $$a, b, C$$, so once again we want $$a \cos C - b = 0$$; it is just more tedious to work out the derivative in the first place, and there are more opportunities for error.

• Thank you for that. I will definitely need to go check what I did but I like this simpler approach instead of using law of cosines. Commented Jul 18 at 10:12
• One question though, where did you find the relation $\tan(B) = \frac{b\sin(C)}{a-b\cos(C)}$? When I look up law of tangents, I get $\frac{a-b}{a+b} = \frac{\tan(\frac{A-B}{2})}{\tan(\frac{A+B}{2})}$. I am not sure how to get your result out of this. Commented Jul 18 at 12:01
• I dropped a perpendicular from point $A$ to side $a$, forming two right triangles. The right triangle on the left has base $a - b\cos C$ and height $b\sin C$ with angle $B$ opposite the side $b\sin C$. Commented Jul 18 at 14:05