Find the areas of the sectors formed by $\angle A C B$ . $\newline$ Give the exact answers in terms of $\pi$ . Do not approximate the answers. $\newline$ Area of small sector $=$ $\square$ $\mathrm{ft}^{2}$ $\newline$ Area of large sector $=$ $\square$ $\mathrm{ft}^{2}$

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Algebra 2

Solve trigonometric equations II

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Q. Find the areas of the sectors formed by

\angle A C B

\newline

Give the exact answers in terms of

\pi

. Do not approximate the answers.

\newline

Area of small sector

=

\square

\mathrm{ft}^{2}

\newline

Area of large sector

=

\square

\mathrm{ft}^{2}

Introduction: To find the area of a sector, we need the central angle in radians and the radius of the circle. The formula is $(\theta/2\pi) * \pi r^2$ , where $\theta$ is the central angle in radians and $r$ is the radius.
Small Sector Area: First, let's find the area of the small sector. Assume the central angle for the small sector is $\theta_1$ and the radius of the circle is $r$ .
Large Sector Area: Area of small sector = $(\theta_1/2\pi) \times \pi r^2$ . We need to plug in the values for $\theta_1$ and $r$ to calculate the area.
Calculation Error: Let's say $\theta_1$ is $60$ degrees, which is $(\pi/3)$ radians. If the radius $r$ is $10$ ft, we substitute these values into the formula.
Calculation Error: Let's say $\theta_1$ is $60$ degrees, which is $(\pi/3)$ radians. If the radius $r$ is $10$ ft, we substitute these values into the formula. Area of small sector = $((\pi/3)/2\pi) * \pi(10)^2 = (1/6) * \pi * 100 = 50\pi/3$ ft$^$2$.
Calculation Error: Let's say $\theta_1$ is $60$ degrees, which is $(\pi/3)$ radians. If the radius $r$ is $10$ ft, we substitute these values into the formula. Area of small sector = $((\pi/3)/2\pi) * \pi(10)^2 = (1/6) * \pi * 100 = 50\pi/3$ ft$^2$. Now, let's find the area of the large sector. Assume the central angle for the large sector is $\theta_2$.
Calculation Error: Let's say $\theta_1$ is $60$ degrees, which is $(\pi/3)$ radians. If the radius $r$ is $10$ ft, we substitute these values into the formula. Area of small sector = $((\pi/3)/2\pi) * \pi(10)^2 = (1/6) * \pi * 100 = 50\pi/3$ ft$^2$. Now, let's find the area of the large sector. Assume the central angle for the large sector is $\theta_2$. The large sector's central angle $\theta_2$ will be the complement of $\theta_1$ in the circle, so $60$$0$.
Calculation Error: Let's say $\theta_1$ is $60$ degrees, which is $\left(\frac{\pi}{3}\right)$ radians. If the radius $r$ is $10$ ft, we substitute these values into the formula. Area of small sector = $\left(\frac{\left(\frac{\pi}{3}\right)}{2\pi}\right) * \pi(10)^2 = \left(\frac{1}{6}\right) * \pi * 100 = \frac{50\pi}{3}$ ft$^2$. Now, let's find the area of the large sector. Assume the central angle for the large sector is $\theta_2$. The large sector's central angle $\theta_2$ will be the complement of $\theta_1$ in the circle, so $60$$0$. $60$$1$ radians.
Calculation Error: Let's say $\theta_1$ is $60$ degrees, which is $\left(\frac{\pi}{3}\right)$ radians. If the radius $r$ is $10$ ft, we substitute these values into the formula. Area of small sector = $\left(\frac{\left(\frac{\pi}{3}\right)}{2\pi}\right) * \pi(10)^2 = \left(\frac{1}{6}\right) * \pi * 100 = \frac{50\pi}{3}$ ft$^2$. Now, let's find the area of the large sector. Assume the central angle for the large sector is $\theta_2$. The large sector's central angle $\theta_2$ will be the complement of $\theta_1$ in the circle, so $60$$0$. $60$$1$ radians. Area of large sector = $60$$2$. We substitute $60$$3$ and $60$$4$ ft into the formula.
Calculation Error: Let's say $\theta_1$ is $60$ degrees, which is $(\pi/3)$ radians. If the radius $r$ is $10$ ft, we substitute these values into the formula. Area of small sector = $((\pi/3)/2\pi) * \pi(10)^2 = (1/6) * \pi * 100 = 50\pi/3$ ft^$2$. Now, let's find the area of the large sector. Assume the central angle for the large sector is $\theta_2$. The large sector's central angle $\theta_2$ will be the complement of $\theta_1$ in the circle, so $\theta_2 = 2\pi - \theta_1$. $60$$0$ radians. Area of large sector = $60$$1$. We substitute $60$$2$ and $60$$3$ ft into the formula. Area of large sector = $60$$4$ ft^$2$.
Calculation Error: Let's say $\theta_1$ is $60$ degrees, which is $\left(\frac{\pi}{3}\right)$ radians. If the radius $r$ is $10$ ft, we substitute these values into the formula. Area of small sector = $\left(\frac{\left(\frac{\pi}{3}\right)}{2\pi}\right) * \pi(10)^2 = \left(\frac{1}{6}\right) * \pi * 100 = \frac{50\pi}{3}$ ft$^2$. Now, let's find the area of the large sector. Assume the central angle for the large sector is $\theta_2$. The large sector's central angle $\theta_2$ will be the complement of $\theta_1$ in the circle, so $60$$0$. $60$$1$ radians. Area of large sector = $60$$2$. We substitute $60$$3$ and $60$$4$ ft into the formula. Area of large sector = $60$$5$ ft$^2$. However, there's a mistake in the calculation. The area of the large sector should be simplified.