4140.11 – Find the Circumference

$\angle B = \angle A = 70^\circ$.

By the law of sines, the radius is:

$$\begin{align*}\dfrac{x}{\sin(70^\circ)} &= \dfrac{6}{\sin(40^\circ)} \\[1em]x &= 6 \left(\dfrac{\sin(70^\circ)}{\sin(40^\circ)}\right)\end{align*}$$

The circumference is $2 \pi x$, or:

$$\begin{align*}2 \pi \times 6 \left(\dfrac{\sin(70^\circ)}{\sin(40^\circ)}\right) &\approx \dfrac{12 (3.14)(.94)}{.64} \\[1em]&\approx 55.34 \ \cm\end{align*}$$

(Note: The diameter of the circle is about $\dfrac{55.34}{3.14}$, or 17.6. The radius is about 8.8, not a whole lot more than that chord.)