1. ## Finding the radius, solving, and finding theta?

#1 In a circle, an arc of length 30 cm contains a central angle of 120 degrees. Determine the radius of this circle.

Work: 30/(2(pi)r) = (3pi/4)/2pi
r= 40/pi... But this isn't right.

#2 Solve: 5(sin(x))^2 = cos(x)

Work: Unsure how to start this and finish it. I also tried graphing it and finding the intersections, but I got HUGE numbers (3 digit range).

#3 The point (5, -6) is on the terminal arm of the standard position angle theta. Determine the smallest positive measure of theta in radians.

Work: graphed coordinates (5, -6), but don't know how to continue.

2. Originally Posted by AlphaRock
#1 In a circle, an arc of length 30 cm contains a central angle of 120 degrees. Determine the radius of this circle.

Work: 30/(2(pi)r) = (3pi/4)/2pi
r= 40/pi... But this isn't right.

#2 Solve: 5(sin(x))^2 = cos(x)

Work: Unsure how to start this and finish it. I also tried graphing it and finding the intersections, but I got HUGE numbers (3 digit range).

#3 The point (5, -6) is on the terminal arm of the standard position angle theta. Determine the smallest positive measure of theta in radians.

Work: graphed coordinates (5, -6), but don't know how to continue.

first question

the angle is 120 in rad it equal $\frac{2\pi}{3}$ this is the angle facing the arc

the arc length given by

$\theta (r)$ theta is the angle facing the arc and r is the radius of the circle so you have theta and the arc length r is unknown

$r(\theta) = arclength$ you solve the equation

second question $5sin^2(x) = cos(x)$

$5(1-cos^2(x) ) = cos(x)$

$5-5cos^2(x) = cos(x) \Rightarrow 5cos^2(x) - cos(x) -5$ let t=cos(x)

$5t^2 - t - 5 = 0$ solve for t by the formula

$\frac{-b\pm \sqrt{b^2-4(a)(b)}}{2a}$ since a,b and c are

$ax^2 + bx + c =0$ after finding t value sub it this t=cosx

the last question the head of the angle in the origin or what ?

3. Originally Posted by AlphaRock
#3 The point (5, -6) is on the terminal arm of the standard position angle theta. Determine the smallest positive measure of theta in radians.

Work: graphed coordinates (5, -6), but don't know how to continue.

\begin{aligned}
Solve for $\theta$ by taking the inverse tangent of both sides. Remember, though, that inverse tangent only gives angles in $\text{-}\frac{\pi}{2} < \theta < \frac{\pi}{2}$. Our point is in Quadrant IV, so we're okay.
$\theta = \tan^{-1}\left(\text{-}\frac{6}{5}\right) \approx \text{-}50.19^{\circ} + 360^{\circ} \approx 309.81^{\circ}$