I am having trouble solving the following highschool textbook exercise:
Let $AM$ and $AN$ be two chords in a circle of radius $r$ having the same length and such that the sine of the acute angle $\widehat{MAN}$ is $11/16$. Divide the arc $MN$ not containing $A$ in three smaller arcs of the same length, $MP,PQ,QM$. Find the perimeter of the quadrilateral $MNQP$. (According to the book the solution is $23/8 r$).
I tried computing the length of the segment $MP$, but that requires a formula for $\sin(x/3)$.
Using the formula $$\sin 3x=3 \sin x -4 \sin^3 x$$ will give a straight forward solution.
$$\sin 3x=\frac{11}{16} $$ $$\implies 3 \sin x -4 \sin^3 x=\frac{11}{16}$$ $$\implies 64 \sin^3x-48\sin x +11=0$$ $$\implies (4 \sin x -1)(16 \sin^2 x +4 \sin x -11)=0$$ $$\implies \sin x = \frac{1}{4}$$
(Note that other solutions for $\sin x$ are rejected because $3x$ has to be an acute angle.)
In the figure, $$\frac{MN}{\sin 3x}= 2r $$
$$\implies MN=2r \times \sin 3x$$ $$\implies MN= 2r \times \frac{11}{16} $$ $$\implies MN= \frac{11r}{8} $$
In $\Delta MNQ$, $$\frac{NQ}{\sin x}=\frac{MN}{\sin (180^\text{0}-3x)}$$ $$\implies NQ= \sin x \times \frac{MN}{\sin 3x}$$ $$\implies NQ= \frac{1}{4} \times \frac{\frac{11r}{8}}{\frac{11}{16}}$$ $$\implies NQ=\frac{r}{2}$$
Therefore the perimeter of $MNQP$ is
$$MN+3NQ=\frac{11r}{8}+3 \times \frac{r}{2}=\frac{23r}{8}$$