Points $A, B, C$ and $D$ are four consecutive vertices of a regular nonagon with sides 25mm long. Find the length of $AD$.
Here's what I have so far:
I believe I've taken the right steps, but I'm not sure how to proceed. I don't think we can solve for $AE$ or $ED$ using the trigonometric ratios as we don't have enough info (or do we?). How do I finish this problem?
Your method would lead to the right solution, but it is unnecessarily complicated. We solve the problem in two steps.
Both the blue and black triangles are right-angled. The blue angle is 20°, so the circumradius (black hypotenuse; distance from any vertex to centre of nonagon) is $(25/2)/\sin20^\circ=36.547\text{ mm}$. The black angle is 60°, so half the chord length (half of $AD$) is $36.547\sin60^\circ=31.651\text{ mm}$. Then $AD$ is twice this, or $63.302\text{ mm}$.