auto generate barcode vb net Fig. 11.9 in Visual Studio .NET

Encoder QR in Visual Studio .NET Fig. 11.9

Fig. 11.9
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In Fig. 11.9, BC represents the tower, DB represents the cliff, and A is the point on the opposite shore. In triangle ABC: / ACB 90 28 40 61 20 / CBA / BAC c 90 180 18 20 (/ CBA 108 20 / ACB) 125 sin 61 20r sin 10 20r 10 20 125(0.8774) 0.1794 611
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a sin / ACB sin / BAC
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CHAPTER 11 Oblique Triangles
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In right triangle ABD: DB AD c sin 18 20 c cos 18 20 611(0.3145) 611(0.9492) 192 580
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The river is 580 ft wide, and the cliff is 192 ft high.
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Case III 11.9 Solve the triangle ABC, given c 628, b 480, and C 55 10 . Refer to Fig. 11.10(a).
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Since C is acute and c > b, there is only one solution. For B: sin B b sin C c 480 sin 55 10r 628 480(0.8208) 628 0.6274 and B 38 50r
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(NOTE: If sin B 0.6274, then B 38 50 or B 180 38 50 141 10 and each could be an angle in a triangle. Since A B C 180 , it follows that C B < 180 ; thus B 141 10 is not a solution in this problem because C B 55 10 141 10 196 20 > 180 . Whenever 0 < sin x < 1, it is possible to find angles x in quadrants I and II that satisfy the value of sin x and could be angles in a triangle. The first-quadrant angle is always a solution but the second-quadrant angle is a solution only when its sum with the given angle is less than 180 .) For A: A 180 b sin A sin B (B C) 86 0 480(0.9976) 0.6271 86 0 , and a 764 764.
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For a: a
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480 sin 86 0r sin 38 50r 38 50 , A
The required parts are B
Fig. 11.10
11.10 Solve the triangle ABC, given a
525, c
421, and A
130 50 . Refer to Fig. 11.10(b).
Since A is obtuse and a > c, there is one solution. For C: For B: For b: sin C B b 180 a sin B sin A c sin A a (C 421 sin 130 50r 525 A) 11 50 525(0.2051) 0.7566 11 50 , and b 142 142. 421(0.7566) 525 0.6067 and C 37 20r
525 sin 11 50r sin 130 50r 37 20 , B
The required parts are C
CHAPTER 11 Oblique Triangles
11.11 Solve the triangle ABC, given a
31.5, b
51.8, and A
51.8(0.5544) 31.5 180 For C : C For c : cr 65 40 180
33 40 . Refer to Fig. 11.11(a).
Since A is acute and a < b, there is the possibility of two solutions. For B: sin B b sin A a (A 51.8 sin 33 40r 31.5 65 40 and B B) 80 40 0.9117 114 20 . (A B) 32 0 31.5 sin 32 0r sin 33 40r 30.1
There are two solutions, B For C: C For c: c 180 a sin C sin A
31.5 sin 80 40r sin 33 40r 56.1
a sin Cr sin A
31.5(0.9868) 0.5544 The required parts are
31.5(0.5299) 0.5544 65 40 , C 114 20 , C 80 40 , and c 32 0 , and c
for triangle ABC: B and for triangle ABC : B
56.1 30.1
Fig. 11.11
11.12 A pilot wishes a course 15 0 against a wind of 25 mi/h from 160 30 . Find his required heading and the groundspeed when the airspeed is 175 mi/h. Refer to Fig. 11.11(b).
Since / BAC is acute and a > c, there is one solution. sin C B b c sin /BAC a 180 a sin B sin / BAC 25 sin 34 30r 175 / ACB) 175 sin 140 50r sin 34 30r 25(0.5664) 175 140 50r 175(0.6316) 0.5664 0.0809 and / ACB 4 40r
(/BAC
Case IV 11.13 Solve the triangle ABC, given a 132, b 224, and C 28 40 . Refer to Fig. 11.12(a).
Fig. 11.12
CHAPTER 11 Oblique Triangles
For c: c2 a2 (132) (132) For A: For B: sin A sin B b2
2ab cos C (224)2 (224)2 2(132)(224) cos 28 40 2(132)(224)(0.8774) 15,714 and c and and 125 A B 30 30 120 40 132(0.4797) 125 224(0.4797) 125
a sin C c b sin C c
132 sin 28 40r 125 224 sin 28 40r 125
0.5066 0.8596
(Since b > a, A is acute; since A Check: A B C
C < 90 , B > 90 .) 30 30 , B 120 40 , and c 125.
179 50 . The required parts are A
11.14 Solve the triangle ABC, given a
For b: b
322, c
212, and B
110 50 . Refer to Fig. 11.12(b).
110 50 ) and and and A C cos 69 10 ] b 444 42 40 26 30
2ca cos B (322)
[cos 110 50
cos (180
(212) For A: For C: Check: sin A sin C A B
a sin B b c sin B b C
2(212)(322)( 0.3557) 197,191 322(0.9346) 322 sin 110 50r 0.6778 444 444 212 sin 110 50r 444 212(0.9346) 444 444. 0.4463
180 . 42 40 , C 26 30 , and b
The required parts are A
11.15 Two forces of 17.5 and 22.5 lb act on a body. If their directions make an angle of 50 10 with each other, find the magnitude of their resultant and the angle that it makes with the larger force.
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