1.
Item 1. Which of the following shows an angle of \(215^\circ\) in standard position?
Solution:An angle in standard position starts on the positive \(x\)-axis, which is the initial arm.Positive angles rotate counterclockwise.\(215^\circ\) is greater than \(180^\circ\) and less than \(270^\circ\).Therefore, the terminal arm of the angle is in Quadrant III.The graph showing an angle of \(215^\circ\) in Quadrant III is Graph C.Answer: C
2.
Item 4. Angle \(\theta\) lies in Quadrant II and \(\cos\theta=-\frac{2}{3}\). Find the exact values of the other two trigonometric ratios.
Solution:Use the ratios:\(\sin\theta=\frac{y}{r}\), \(\cos\theta=\frac{x}{r}\), and \(\tan\theta=\frac{y}{x}\).Given \(\cos\theta=-\frac{2}{3}\), we have \(x=-2\) and \(r=3\).Since \(\theta\) is in Quadrant II, \(y\) is positive.Find \(y\) using the Pythagorean Theorem:\(x^2+y^2=r^2\)\((-2)^2+y^2=3^2\)\(4+y^2=9\)\(y^2=5\)\(y=\sqrt5\)Now find the other two ratios:\(\sin\theta=\frac{y}{r}=\frac{\sqrt5}{3}\)\(\tan\theta=\frac{y}{x}=\frac{\sqrt5}{-2}=-\frac{\sqrt5}{2}\)Answer: C
3.
Item 2. Determine the measures of three angles greater than \(45^\circ\) and less than \(360^\circ\) that have a reference angle of \(45^\circ\).
Solution:A reference angle of \(45^\circ\) means the acute angle formed with the \(x\)-axis is \(45^\circ\).In Quadrant I, the angle is \(45^\circ\), but the question asks for angles greater than \(45^\circ\), so we do not include it.In Quadrant II:\(180^\circ-45^\circ=135^\circ\)In Quadrant III:\(180^\circ+45^\circ=225^\circ\)In Quadrant IV:\(360^\circ-45^\circ=315^\circ\)Therefore, the three angles are \(135^\circ,225^\circ,315^\circ\).Answer: B
4.
Item 5. Determine the exact value of \(\sin 60^\circ\).
Solution:Use the special \(30^\circ-60^\circ-90^\circ\) triangle.In this triangle, the side opposite \(60^\circ\) is \(\sqrt3\), and the hypotenuse is \(2\).Therefore:\(\sin60^\circ=\frac{\text{opposite}}{\text{hypotenuse}}\)\(\sin60^\circ=\frac{\sqrt3}{2}\)Answer: D
5.
Item 6. Determine the exact value of \(\cos(-120^\circ)\).
Solution:Draw \(-120^\circ\) in standard position. A negative angle rotates clockwise.The angle \(-120^\circ\) lands in Quadrant III.The reference angle is:\(180^\circ-120^\circ=60^\circ\)From the special triangle:\(\cos60^\circ=\frac{1}{2}\)Cosine is negative in Quadrant III.Therefore:\(\cos(-120^\circ)=-\frac{1}{2}\)Answer: D
6.
Item 3. Point \(P(1,4)\) is on the terminal arm of angle \(\theta\) in standard position. Calculate \(\sin\theta\).
Solution:The point is \(P(1,4)\), so \(x=1\) and \(y=4\).Find the radius \(r\) using the Pythagorean Theorem:\(r=\sqrt{x^2+y^2}\)\(r=\sqrt{1^2+4^2}\)\(r=\sqrt{1+16}\)\(r=\sqrt{17}\)Use the sine ratio:\(\sin\theta=\frac{y}{r}\)\(\sin\theta=\frac{4}{\sqrt{17}}\)Rationalize the denominator:\(\frac{4}{\sqrt{17}}\cdot\frac{\sqrt{17}}{\sqrt{17}}=\frac{4\sqrt{17}}{17}\)Answer: B
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