Math141Sols3S11

Math141Sols3S11 - MIDTERM 3 SOLUTIONS (CHAPTER 4) MATH 141...

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MIDTERM 3 SOLUTIONS (CHAPTER 4) MATH 141 – SPRING 2011 – KUNIYUKI 150 POINTS TOTAL: 35 FOR PART 1, AND 115 FOR PART 2 Show all work, simplify as appropriate, and use “good form and procedure” (as in class). Box in your final answers! No notes or books allowed. PART 1: SCIENTIFIC CALCULATORS ALLOWED! (35 POINTS TOTAL) Write units in your final answers where appropriate. Try to avoid rounding intermediate results; if you do round off, do it to at least five significant digits. 1) A central angle of a circle has measure 0.5 radians. It intercepts an arc of length 12 centimeters along the circle. What is the radius of the circle? (4 points) Arc length s = r ± 12 = r 0.5 () 12 0.5 = r r = 24 cm 2) Give the solutions for sin = ² 0.8 , where 0 ± ² < 2 ³ . Give your answers in radians, and round them off to the nearest thousandth of a radian (i.e., to three decimal places). (8 points) In radians (the assumed measure), sin ± 1 ± 0.8 ²± 0.927 , a Quadrant IV angle. This radian measure is not in the specified interval, 0, 2 ² ³ ) , so we will add 2 . One of our solutions is: sin ± 1 ± 0.8 + 2 ³ 5.356 . The only other Quadrant in which sin is negative in value is Quadrant III. We also want the brother in Quadrant III that is in the interval 0, 2 ² ³ ) . Remember that “brothers” have the same reference angle; in this case, it is sin ± 1 0.8 . The desired brother is given by: sin ± 1 0.8 + ³ 4.069 .
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(Intersection points have the same y -coordinate, ± 0.8 .) [Approximate] Solution set: 4.069, 5.356 {} 3) Write sec sin ± 1 x 3 ² ³ ´ µ · ² ³ ´ µ · as an equivalent algebraic expression, as we have done in class. Assume x is in the domain of the expression. (7 points) Let ± = sin ² 1 x 3 ³ ´ µ · ¸ . Then, sin = x 3 . We may assume is acute when we draw our right triangle model using SOH-CAH-TOA. We use the Pythagorean Theorem to find the missing (blue) side length, b : x () 2 + b 2 = 3 2 x 2 + b 2 = 9 b 2 = 9 ± x 2 b = ± x 2 Take "+" root. By SOH-CAH-TOA, cos = Adj. Hyp. , so its reciprocal, sec = Hyp. Adj. . sec sin ± 1 x 3 ² ³ ´ µ · ² ³ ´ µ · = sec ¸ = 3 ± x 2 Warning: 9 ± x 2 ¹ 3 ± x , usually = 39 ± x 2 9 ± x 2
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4) An airplane is flying at an altitude of 10,000 feet over a flat desert. In other
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Math141Sols3S11 - MIDTERM 3 SOLUTIONS (CHAPTER 4) MATH 141...

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