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MATH108DE Practice 1 Problems 5 problems SOLUTIONS text: College Algebra with Trigonometry, 7th ed by Barnett, Ziegler, Byleen McGraw-Hill Practice 1 Problems sec6.1 ....page 421, prob 14, 20 sec6.2 ....page 433, prob 26 sec6.3 ....page 442, prob 12 sec6.4 ....page 453, prob 64 ______________________________________________________________________ Sec 6.1 page 421, prob 14 Converting from degree measure of an angle to radian measure of an angle. SOLUTION Since rad 180 to get the conversion factor to convert from degrees to radians we simply divide the above equation by 180 to get 1 rad . 180 So 60 60 rad 180 120 120 180 180 3 2 rad 180 3 rad 180 240 240 4 rad 180 3 5 rad 300 300 180 3 Sec6.1 page 421, prob 20 Converting from radian measure of an angle to degree measure of an angle. SOLUTION The angles in this problem are all negative angles. That means that the angle is generated by a clockwise rotation of the terminal side of the angle. A positive angle is generated by a counterclockwise rotation of the terminal side of the angle. Sine rad 180 to get the conversion factor to convert from radians to degrees we simply divide the above equation by to get 1 rad 180 . However, in this problem all the angles are expressed in terms of , so the arithmetic is easier. rad 180 45 4 4 180 rad 90 2 2 3 3 180 rad 135 4 4 rad 1180 180 Sec6.2 page 433, prob 26 We have to solve a right triangle. That means we must find the measures of all the sides and all the angles of the right triangle. We are given the data that c 42.5 6230 SOLUTION First, because it's easiest to do, let's find the measure of the complementary angle 90 90 6230 8960 6230 2730 The measure of the side c is given. To find the measures of the other two sides, we can use the sine function and the cosine function. b c sin 42.5 sin 6230 42.50.88701 37.7 The value is rounded to one decimal place. a c cos 42.5 cos6230 42.5 0.46175 19.6 The value is rounded to one decimal place. Sec6.3 page 442, prob 12 Find the values of the other five trig functions given that csc 5 ; 4 is an angle in quadrant III. SOLUTION The sign of a trig function in each of the four quadrants can be read from the signs of the coordinates of a point a, b in each of the four quadrants. Quadrant II Quadrant I Quadrant III Quadrant IV Remember that r is the positive distance of the point from the origin of the coordinate axes, 0,0 . An angle is said to be in a certain quadrant when the terminal side of the angle is in that quadrant. For this problem the angle is in quadrant III. That means that the terminal side of the angle in quadrant III. A point in quadrant III has an x-coordinate that is negative and a y-coordinate that also is negative. That is, a point will have coordinates ( - , -). csc 5 4 Use the reference angle associated with the angle . NOTE: The above figure is drawn for a point P in quadrant IV. For our problem the point P(a,b) is in quadrant III. 5 5 r csc 4 4 b and a 5 4 25 16 9 3 2 2 So the point P in the reference triangle has the coordinates P( -3, -4 ). Using the definitions of the trig functions for any angle, we get 4 4 5 5 3 3 cos 5 5 5 5 sec 3 3 4 4 tan 3 3 3 3 cot 4 4 sin Sec6.4 page 453, prob 64 Find the smallest positive angle csc 2 . in degree measure for which SOLUTION From the definition of the cosecant function for any angle csc r , b since r is always positive, the denominator must be negative in order for the fraction itself to be negative, csc 2 2 . 1 That is, r = 2 and b= -1. So the terminal side of the angle must be in quadrant III and 2 2 a 2 1 3 . The reference triangle is then the 30-60 right triangle From the above figure 2 csc30 2 1 is 30 and the smallest value for - remember is an angle in quadrant III - would then be so the reference angle associated with 180 30 210 is the final answer. - END