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60. First note that Fig. 63 contains 7 node points (including the zero-volt reference node at ground potential), but note that only two of the node voltages, at nodes 1 and 2, are unknown. Let us therefore begin by writing the Kirchho current law equations for nodes 1 and 2; if we take the current direction as they happen to be drawn in Fig. 63 we have at node 1: I1 I2 I3 I4 0 A at node 2: I4 I5 I6 0 B
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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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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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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:
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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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66. Let us make use of the right triangle of Fig. 64 and eqs. (69) and (70) in section 5.2, as follows.
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