Question

find one positive and one negative angle coterminal with an angle of $\frac{9\pi}{6}$. $\frac{11\pi}{6};\frac{-7\pi}{6}$ $\frac{3\pi}{6};\frac{-21\pi}{6}$ $\frac{21\pi}{6};\frac{-3\pi}{6}$ $\frac{7\pi}{6};\frac{-11\pi}{6}$ question 6 (5 points) find the area of a sector with a central angle of $170^{\circ}$ and a radius of 17 millimeters. round to the nearest tenth. $857.5\text{ mm}^2$ $100.9\text{ mm}^2$ $428.7\text{ mm}^2$ $25.2\text{ mm}^2$

Explanation:

Step1: Formula for coterminal angles

Coterminal angles are found by adding or subtracting \(2\pi\) (or \(\frac{12\pi}{6}\)) from the given angle \(\frac{9\pi}{6}\).
For positive coterminal angle: \(\frac{9\pi}{6}+\frac{12\pi}{6}=\frac{21\pi}{6}\)
For negative coterminal angle: \(\frac{9\pi}{6}-\frac{12\pi}{6}=\frac{- 3\pi}{6}\)

Step2: Formula for area of a sector

The formula for the area of a sector is \(A=\frac{\theta}{360}\times\pi r^{2}\), where \(\theta = 170^{\circ}\) and \(r = 17\) mm.
Substitute the values: \(A=\frac{170}{360}\times\pi\times(17)^{2}\)
\(A=\frac{170}{360}\times\pi\times289\)
\(A=\frac{170\times289\pi}{360}\)
\(A=\frac{49130\pi}{360}\approx428.7\) \(mm^{2}\)

Answer:

For the first question: \(\frac{21\pi}{6};\frac{-3\pi}{6}\) (third option)
For the second question: \(428.7\) \(mm^{2}\) (third option)