Find the area of the sector inside the square ABCD. Dimensions shown are in centimeter. A. 52.36 cm2 C. 48.32 cm2 B. 64.36 cm2 D. 32.94 cm2 <strong>Solution</strong… Jhun Vert Wed, 06/12/2024 - 20:30 Solution Click here to expand or collapse this section $\cos \alpha = 5/10$ $\alpha = 60^\circ$ $\begin{align} \theta & = 180^\circ - 2\alpha \\ \\ & = 180^\circ - 2(60^\circ) \\ \\ & = 60^\circ \end{align}$ $\begin{align} A & = \dfrac{\pi r^2 \theta_{deg}}{360^\circ} \\ \\ & = \dfrac{\pi(10^2)(60^\circ)}{360^\circ} \\ \\ & = 52.36 ~ \text{cm}^2 \end{align}$ Log in or register to post comments Log in or register to post comments
<strong>Solution</strong… Jhun Vert Wed, 06/12/2024 - 20:30 Solution Click here to expand or collapse this section $\cos \alpha = 5/10$ $\alpha = 60^\circ$ $\begin{align} \theta & = 180^\circ - 2\alpha \\ \\ & = 180^\circ - 2(60^\circ) \\ \\ & = 60^\circ \end{align}$ $\begin{align} A & = \dfrac{\pi r^2 \theta_{deg}}{360^\circ} \\ \\ & = \dfrac{\pi(10^2)(60^\circ)}{360^\circ} \\ \\ & = 52.36 ~ \text{cm}^2 \end{align}$ Log in or register to post comments
<strong>Solution</strong…