Respuesta :

LammettHash

a. Your answer and reasoning are correct.

b. If O is the center of the circle, then angle ABO has measure 72º, so that by the law of cosines

[tex]AB^2=6^2+6^2-2\cdot6\cdot6\cos72^\circ\implies AB=3\sqrt{10-2\sqrt5}[/tex]

just to add to the superb reply above by @LammettHash

[tex]\bf \textit{arc's length}\\\\ s=\cfrac{\pi \theta r}{180}~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ \cline{1-1} r=6\\ \theta =72 \end{cases}\implies s=\cfrac{\pi (72)(6)}{180}\implies s=\cfrac{12\pi }{5}\implies s\approx 7.54[/tex]

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