Originally Posted by

**Grandad** Hello simplependulumInteresting indeed!

But just slightly flawed, I'm afraid, because it's impossible to draw so that $\displaystyle \angle BCD = \frac{\pi}{6}$.

The fact is that $\displaystyle x < 0$, and your solution of $\displaystyle x^2 = \frac{3}{4(13)} $ stops short of solving for $\displaystyle x$. If you take the positive square root and substitute back, you'll find that $\displaystyle \angle BCD \approx 76.1^o$ and $\displaystyle \angle ACD \approx 73.9^o$. This gives $\displaystyle \angle BCD = 150^o$, which represents the solution to the equation $\displaystyle \arcsin 4x + \arccos x = \frac{5\pi}{6}$.

The negative solution to your equation is, of course, the correct one - and it's the same as the one I obtained by a more conventional approach. Sadly, it's not possible to draw your diagram with a negative value of $\displaystyle x$.

Unless, of course, you can think of a modification ...?

Grandad