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Now apply eq. (68) to each current at nodes 1 and 2, then substitute into (A) and (B). Doing this (paying careful attention to battery polarities), eqs. (A) and (B) become 8 V1 V1 6 V1 V1 V2 0 6 15 5 8 and V1 V2 V2 12 10 V2 0 8 7 9 Now multiply the rst equation by 240 and the second by 504; this should give you the two simultaneous equations 67V1 15V2 304 63V1 191V2 48 the solutions of which are (using determinants is probably easiest) V1 4:9598 volts; answer; and V2 1:8873 volts; answer; both voltages with respect to the zero-volt reference node. 61. Since the trigonometric functions are de ned in terms of the ratios of the lengths of the sides of a right triangle to one another, let us, for simplicity, use the triangle below.

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(Standalone mode means your GUI must access the database without going through the network server. In other words, without RMI or sockets.) Passing arguments to the command-line of your main class looks like this:

p (By the Pythagorean theorem, the length of the vertical dashed line is equal to 3, as shown.) The answers are now found as follows. First, if we take the 608 angle as the reference angle, we have (see the de nitions just prior to eq. 69) p sin 608 3=2 0:866025 approx: and cos 608 1=2 0:500000

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So anything you type after the JAR name is a command-line argument that gets passed straight on through to the main() method of your main class. In the example above, the ExamProject main() method would get the string server at args[0]. Running javadoc on Your Package 16 describes the details of javadoc, but we thought a little reminder about running javadoc on your package might be helpful here. For this example, assume that

Your class files are in the suncertify/db directory. Your working directory is one level above the start of the package (in other

The zero is below the frequency of the output lter pole in order to improve phase margin. If the output lter has a lower Q (it normally does not), then the zero could be at a frequency that is closer to the output lter pole. An additional zero exists, because of the output lter capacitor ESR at F= 1 = 14, 475 Hz 2 R1 C7

Next, taking the 308 angle as the reference angle, we have sin 308 1=2 0:500000 and cos 308 p 3=2 0:866025; approx:

words, one level above the suncertify directory).

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To create javadocs for all classes in the suncertify.db package and have them placed in the docs/javadocs directory (where they ll need to be in our sample packaging instructions), run the following command from the directory one level above both docs and suncertify:

62. First, by de nition, sin b=h (using as reference angle); also, by de nition, cos b=h (using as reference angle). Thus, sin cos , or, since 90 , we have sin cos 908 ; answer: 63. From the answer to problem 62 we have that sin 62838 0 cos 908 62838 0 . Thus the required angle is 908 62838 0 , and hence (note that 908 89860 0 we have 89860 0 62838 0 27822 0 ; answer: 64. (a) In the right triangle of Fig. 65, by the theorem of Pythagoras, a2 b2 h2 . However, by eqs. (69) and (70), a h cos and b h sin . Making these substitutions gives the required identity. Using eq. (71) and then eqs. (69) and (70), we have b h sin sin tan ; as required: a h cos cos 65. (a) 115 degrees is a second quadrant angle, hence (Fig. 76) 180 115 65; thus cos 115 cos 65 0:4226; answer: (b) (c) (d) (e) (f) By eq. (76), sin 35 sin 35 0:5736; answer: From Fig. 76, tan 155 tan 25 0:4663; answer: 255 degrees is a third quadrant angle, hence (Fig. 77) 255 180 75; thus sin 255 sin 75 0:9659; answer: From Fig. 76, cos 95 cos 85 0:0872; answer: By eq. (78), tan 285 tan 285. Since 285 degrees is a fourth quadrant angle, we have (Fig. 78), tan 285 tan 75 3:7321; answer: (g) (h) sin 285 sin 75 0:9659; answer: By eq. (76), sin 188 sin 188. Since 188 is a third quadrant angle, we have (Fig. 77) sin 188 sin 8 0:1392; answer:

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