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Item 1. Formula Sheet-PC Math 12. The formula sheet is given for this course, as found in the course introduction.
This item is a formula sheet/reference item. No calculation is required. Use the formula sheet as needed throughout the test. Answer: N/N
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Item 2. Angie is observing a squirrel on a tree branch. If the angle of elevation is \(25^\circ\) and Angie judges that the distance between her and the squirrel is \(20\text{ m}\), how far off the ground is the squirrel? Give your answer to the nearest tenth of a meter.
Use the sine ratio because the opposite side and hypotenuse are involved. \(\sin A=\frac{\text{opposite}}{\text{hypotenuse}}\) \(\sin25^\circ=\frac{\text{height}}{20}\) \(\text{height}=20\sin25^\circ\) \(\text{height}\approx20(0.4226)\) \(\text{height}\approx8.452\text{ m}\) Rounded to the nearest tenth, the height is \(8.5\text{ m}\). Answer: D
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Item 3. Find \(\theta\) if \(x=24\) and \(z=9\).
From the diagram, \(z\) is adjacent to \(\theta\), and \(x\) is the hypotenuse. Use the cosine ratio. \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\) \(\cos\theta=\frac{z}{x}=\frac{9}{24}\) \(\theta=\cos^{-1}\left(\frac{9}{24}\right)\) \(\theta\approx67.98^\circ\) Rounded to the nearest degree, \(\theta\approx68^\circ\). The provided answer key marks this item as D. Answer: D
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Item 4. Determine the length of \(AC\) to the nearest tenth of a centimetre.
To determine \(AC\), find \(AB\) and \(BC\), then add them together. For the left triangle, use cosine: \(\cos64^\circ=\frac{AB}{38.9}\) \(AB=38.9\cos64^\circ\approx17.05\text{ cm}\) For the right triangle, use cosine: \(\cos55^\circ=\frac{BC}{43.3}\) \(BC=43.3\cos55^\circ\approx24.84\text{ cm}\) \(AC=AB+BC\) \(AC\approx17.05+24.84=41.89\text{ cm}\) Rounded to the nearest tenth, \(AC\approx41.9\text{ cm}\). Answer: D
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Item 5. Determine the measures of three angles greater than \(45^\circ\) and less than \(360^\circ\) that have a reference angle of \(45^\circ\).
A reference angle is the positive acute angle between the terminal arm and the \(x\)-axis. For a reference angle of \(45^\circ\): In Quadrant II, \(\theta=180^\circ-45^\circ=135^\circ\). In Quadrant III, \(\theta=180^\circ+45^\circ=225^\circ\). In Quadrant IV, \(\theta=360^\circ-45^\circ=315^\circ\). Therefore the three angles are \(135^\circ,225^\circ,315^\circ\). Answer: A
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Item 6. Which of the following shows an angle of \(315^\circ\) in standard position? [Graphs will be inserted later.]
An angle in standard position starts on the positive \(x\)-axis. A positive angle rotates counterclockwise. \(315^\circ\) is greater than \(270^\circ\) and less than \(360^\circ\), so the terminal arm is in Quadrant IV. The graph that shows this angle is Graph C. Answer: C
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Item 7. Convert \(4.2\) radians to degrees to 1 decimal place.
Convert radians to degrees using \(\pi\text{ radians}=180^\circ\). \(4.2\text{ radians}\cdot\frac{180^\circ}{\pi}\approx240.6^\circ\) Therefore \(4.2\) radians is approximately \(240.6^\circ\). Answer: D
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Item 8. The radian measure of an angle is \(\frac{3\pi}{4}\). The arc that subtends the angle has a length of \(6\pi\). Determine the radius of the circle.
Use the arc length formula \(a=r\theta\). Here \(a=6\pi\) and \(\theta=\frac{3\pi}{4}\). \(6\pi=r\left(\frac{3\pi}{4}\right)\) \(r=\frac{6\pi}{\frac{3\pi}{4}}\) \(r=6\pi\cdot\frac{4}{3\pi}=8\). Answer: C
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Item 9. Which pair of angles are co-terminal with \(\frac{7\pi}{4}\) radians?
Coterminal angles differ by integer multiples of \(2\pi\). Start with \(\frac{7\pi}{4}\). \(\frac{7\pi}{4}+2\pi=\frac{7\pi}{4}+\frac{8\pi}{4}=\frac{15\pi}{4}\) \(\frac{7\pi}{4}-2\pi=\frac{7\pi}{4}-\frac{8\pi}{4}=-\frac{\pi}{4}\) Therefore the coterminal pair is \(-\frac{\pi}{4}\) and \(\frac{15\pi}{4}\). Answer: B
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Item 10. Determine the exact value of \(\sin\left(\frac{\pi}{2}\right)+\cos\pi\).
Evaluate each trigonometric value separately. \(\sin\left(\frac{\pi}{2}\right)=1\) \(\cos\pi=-1\) \(\sin\left(\frac{\pi}{2}\right)+\cos\pi=1+(-1)=0\). Answer: C
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Item 11. Determine the approximate values of \(\sin\theta\) and \(\cos\theta\) from the diagram below. [Diagram will be inserted later.]
On the unit circle, \(x=\cos\theta\) and \(y=\sin\theta\). From the diagram, the point is approximately \((0.198,-0.980)\). Therefore \(\cos\theta\approx0.198\) and \(\sin\theta\approx-0.980\). Answer: B
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Item 12. Visualize the graph of \(\sin\theta\) for \(-2\pi\leq\theta\leq2\pi\). For what values of \(\theta\) is the graph at the maximum value?
The maximum value of \(\sin\theta\) is \(1\). On the interval \(-2\pi\leq\theta\leq2\pi\), \(\sin\theta=1\) at: \(\theta=-\frac{3\pi}{2}\) and \(\theta=\frac{\pi}{2}\). Answer: D
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Item 13. Determine the period of the sinusoidal graph below.
To determine the period, find the horizontal distance between two consecutive matching points on the sinusoidal graph. Tracing one complete cycle shows that the graph repeats every \(\frac{\pi}{2}\) units. Therefore the period is \(\frac{\pi}{2}\). Answer: A
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Item 14. Visualize the graph of \(\cos\theta\) for \(-2\pi\leq\theta\leq2\pi\). For what values of \(\theta\) is the graph at the maximum value?
The maximum value of \(\cos\theta\) is \(1\). On the interval \(-2\pi\leq\theta\leq2\pi\), \(\cos\theta=1\) at: \(\theta=-2\pi,0,2\pi\). Answer: D
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Item 15. A sinusoidal function of the form \(y=a\sin(x-c)+d\) is changed by decreasing \(c\) by \(5\). The new graph will have transformed:
In the form \(y=a\sin(x-c)+d\), the value \(c\) represents the horizontal translation or phase shift. Decreasing \(c\) by \(5\) shifts the graph horizontally \(5\) units to the left. Therefore the new graph is transformed \(5\) units left. Answer: A
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Item 16. The graph of a periodic trigonometric function of the form \(y=a\sin(x-c)+d\) is graphed below. Determine the value of \(d\).
In the function \(y=a\sin(x-c)+d\), \(d\) represents the vertical translation or equilibrium line. The equilibrium line is the average of the maximum and minimum values. \(d=\frac{\text{max}+\text{min}}{2}\) From the graph, \(\text{max}=1\) and \(\text{min}=-3\). \(d=\frac{1+(-3)}{2}=-1\). Answer: A
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Item 17. Of the following equations, which have undergone a horizontal contraction from their original equation: \(y=\sin x\) or \(y=\cos x\)? I. \(y=\cos 3x\) II. \(y=\sin\left(\frac{3}{2}x\right)\) III. \(y=\cos\left(\frac{1}{4}x\right)\) IV. \(y=\sin\left(\frac{2}{3}x\right)\) V. \(y=\cos\left(\frac{4}{3}x\right)\)
For \(y=\sin bx\) or \(y=\cos bx\), a horizontal contraction occurs when \(|b|>1\). I. \(b=3>1\), so this is a contraction. II. \(b=\frac{3}{2}>1\), so this is a contraction. III. \(b=\frac{1}{4}<1\), so this is an expansion. IV. \(b=\frac{2}{3}1\), so this is a contraction. Therefore I, II, and V have undergone horizontal contraction. Answer: C
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Item 18. Given \(y=\sin\left(\frac{1}{2}x\right)\), what is the period of this function?
For \(y=\sin nx\), the period is \(\frac{2\pi}{|n|}\). Here \(n=\frac{1}{2}\). \(\text{Period}=\frac{2\pi}{\frac{1}{2}}=4\pi\). Answer: C
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Item 19. For the function \(f(x)=3\sin 4\left(x+\frac{\pi}{4}\right)-1\), the period and the phase shift respectively are:
For \(f(x)=3\sin 4\left(x+\frac{\pi}{4}\right)-1\), the value of \(b\) is \(4\). The period is \(\frac{2\pi}{b}\). \(\text{Period}=\frac{2\pi}{4}=\frac{\pi}{2}\). The expression \(x+\frac{\pi}{4}\) means the graph shifts left by \(\frac{\pi}{4}\). Therefore the period is \(\frac{\pi}{2}\), and the phase shift is \(\frac{\pi}{4}\) left. Answer: C
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Item 20. At a seaport, the depth of the water, \(d\), in metres, at time \(t\) hours, during a certain day is given by: \(d=3.4\sin\left(\frac{2\pi(t-7.00)}{10.6}\right)+2.8\). On that day, determine the depth of the water at 6:30 PM.
Convert \(6{:}30\text{ PM}\) to 24-hour decimal time. \(6\text{ PM}=18\) hours and \(30\) minutes is \(0.5\) hours. So \(t=18.5\). Substitute \(t=18.5\) into the equation. \(d=3.4\sin\left(\frac{2\pi(18.5-7.00)}{10.6}\right)+2.8\) \(d\approx3.4\sin(6.8167)+2.8\) \(d\approx3.4(0.50853)+2.8\) \(d\approx4.53\text{ m}\). Answer: D
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Item 21. What is the period of the function \(y=2\tan\left(\frac{x}{2}+3\right)\)?
The period of a tangent function is \(\frac{\pi}{b}\). Factor out the coefficient of \(x\). \(y=2\tan\left(\frac{x}{2}+3\right)\) \(y=2\tan\left(\frac{1}{2}(x+6)\right)\) So \(b=\frac{1}{2}\). \(\text{Period}=\frac{\pi}{\frac{1}{2}}=2\pi\). Answer: D
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