Fig. 11.13 in .NET framework

Printing QR Code in .NET framework Fig. 11.13

Fig. 11.13
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Refer to Fig. 11.13(a). In the parallelogram ABCD, / DAB In the triangle ABC, b2 c2 a2 2ca cos B (17.5)2
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/ BCD
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180 and B 129 50 ) and b and
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50 10
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129 50 .
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[cos 129 50
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cos (180 1317
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cos 50 10 ] 36.3 / BAC 21 40
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(22.5)2 sin / BAC
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2(22.5)(17.5)( 0.6406)
a sin B b
17.5 sin 129 50r 36.3
17.5(0.7679) 36.3
The resultant is a force of 36.3 lb and the required angle is 21 40 .
CHAPTER 11 Oblique Triangles
11.16 From A a pilot flies 125 km in the direction N38 20 W and turns back. Through an error, the pilot then flies 125 km in the direction S51 40 E. How far and in what direction must the pilot now fly to reach the intended destination A
Refer to Fig. 11.13(b). Denote the turn-back point as B and the final position as C. In the triangle ABC, b2 c2 a2
2ca cos / ABC (125)2 2(125)(125) cos 13 20 843.7 and b 29.0 0.9940 and / BAC 83 40 0.9730)
(125)
2(125) (1 sin / BAC Since / CAN1 from C to A. a sin /ABC b / BAC
125 sin 13 20r 29.0 / N1AB
125(0.2306) 29.0
45 20 , the pilot must fly a course S45 20 W for 29.0 km in going
Case V 11.17 Solve the triangle ABC, given a
For A: For B: For C: cos A cos B cos C B b2 c2 a2 c2 2bc a2 2ca b2 2ab 180 a2 b2 c2
25.2, b
(37.8)2 (43.4)2 (25.2)2
37.8, and c
43.4. Refer to Fig. 11.14(a).
0.8160 and A 0.4982 and B 0.0947 and C 35 20r 60 10r 84 30r
(43.4)2 (25.2)2 2(37.8)(43.4) (25.2)2 (37.8)2 2(43.4)(25.2) (37.8)2 (43.4)2 2(25.2)(37.8)
Check: A
Fig. 11.14
11.18 Solve the triangle ABC, given a
For A: For B: For C: cos A cos B cos C B b2 c2 a2 C c2 2bc a2 2ca b2 2ab 180 a2 b2 c2
30.3, b
(40.4)2 (62.6)2 (30.3)2
40.4, and c
62.6. Refer to Fig. 11.14(b):
0.9159 and A 0.8448 and B 0.5590 and C 23 40r 32 20r 124 0r
(62.6)2 (30.3)2 2(40.4)(62.6) (30.3)2 (40.4)2 2(62.6)(30.3) (40.4)2 (62.6)2 2(30.3)(40.4)
Check: A
CHAPTER 11 Oblique Triangles
11.19 The distances of a point C from two points A and B, which cannot be measured directly, are required. The line CA is continued through A for a distance 175 m to D, the line CB is continued through B for 225 m to E, and the distances AB 300 m, DB 326 m, and DE 488 m are measured. Find AC and BC. See Fig. 11.15.
Triangle ABC may be solved for the required parts after the angles / BAC and / ABC have been found. The first angle is the supplement of / BAD and the second is the supplement of the sum of / ABD and / DBE. In the triangle ABD whose sides are known, cos / BAD and and and BAD cos / ABD ABD (175)2 (300)2 (326)2 2(175)(300) 0.1367
82 10 (300)2 (326)2 (175)2 2(300)(326) 0.8469
32 10
In the triangle BDE whose sides are known, cos /DBE In the triangle ABC: (225)2 (326)2 (488)2 2(225)(326) 180 180 180 and / DBE 97 50 / DBE) / ABC) 24 10 58 0 145
0.5538 / BAD (/ ABD (/ BAC
123 40r
300, / BAC / ABC
and Then AC
/ ACB
AB sin /ABC sin / ACB AB sin / BAC sin / ACB 145 m and BC
300 sin 24 10r sin 58 0r 300 sin 97 50r sin 58 0r 350 m.
300(0.4094) 0.8480 300(0.9907) 0.8480
The required distances are AC
Fig. 11.15
CHAPTER 11 Oblique Triangles
Case II 11.20 Solve the triangle ABC, given a
B 180 (A C) 54.2 . c c c a sin C sin A 38.1 sin 79.3 sin 46.5 51.6 b b b a sin B sin A 38.1 sin 54.2 sin 46.5 42.6
38.1, A
46.5 , and C
74.3 . See Fig. 11.16.
Fig. 11.16
11.21 Solve the triangle ABC, given b
B 180 (C A) 43.85 . a a a
282.7, A
111.72 , and C
24.43 . See Fig. 11.17.
b sin A sin B 282.7 sin 111.72 sin 43.85 379.1
c C c
b sin C sin B 282.7 sin 24.43 sin 43.85 168.8
Fig. 11.17
Case III 11.22 Solve the triangle ABC, given b 67.25, c 56.92, and B 65.27 . See Fig. 11.18.
There could be two solutions in this case. sin C sin C sin C C 50.25 B C C c sin B b 56.92 sin 65.27 67.25 0.7688 180 65.27 50.25 125.75 125.75 195.02 180
CHAPTER 11 Oblique Triangles
Thus, C is not a solution. A a a a The required parts are C 50.25 , A 180 b sin A sin B 67.25 sin 64.48 sin 65.27 66.82 66.82. (B C) 64.48
64.48 , and a
Fig. 11.18
11.23 Solve the triangle ABC, given a
123.2, b
155.4, A
16.57 . See Fig. 11.19.
Fig. 11.19
Since A is acute and a < b, there may be two solutions. sin B sin B sin B B 21.08 A Therefore, B is a solution. C c c c 180 a sin C sin A 123.2 sin 142.35 sin 16.57 263.9 (A B) 142.35 Cr cr cr cr 180 a sin Cr sin A 123.2 sin 4.51 sin 16.57 33.97 (A Br) 4.51 B B b sin A a 155.4 sin 16.57 123.2 0.3597 180 16.57 21.08 158.92 158.92 175.49 < 180
CHAPTER 11 Oblique Triangles
The required parts are: for for ABC: AB C : B B 21.08 , C 142.35 , and c 4.51 , and c 263.9 33.97
158.92 , C
Case IV 11.24 Solve the triangle ABC, given a 2526, c 1388, B 54.42 . See Fig. 11.20.
Fig. 11.20
b2 b
2ac cos B (1388)2 2(2526)(1388) cos 54.42
(2526)
4,227,261.8 2056 cos C cos C cos C C 33.30 180 b2 c2 2ab (2526)2 (2056)2 (1388)2 2(2526)(2056) a2 0.8358 33.30
b cos A cos A cos A A Check: A B C b2
c2 a2 2bc (2056)2 (1388)2 (2526)2 2(2056)(1388)
0.03977 92.28 92.28 54.42
11.25 Solve the triangle ABC, given b
472.1, c
607.4, A
2bc cos A (607.4)2
125.23 . See Fig. 11.21.
2(472.1)(607.4) cos 125.23
a2 a2 a cos B cos B cos B B Check: A B a2
(472.1)2 922,646.52 960.5
c2 b2 2ac (960.5)2 (607.4)2 (472.1)2 2(960.5)(607.4) 0.9158 23.68 C 125.23 23.68 31.11
cos C cos C cos C C 180.02
b2 c2 2ab (960.5)2 (472.1)2 (607.4)2 2(960.5)(472.1) 0.8562 31.11
CHAPTER 11 Oblique Triangles
Fig. 11.21
11.26 Two adjacent sides of a parallelogram are 3473 and 4822 ft, and the angle between them is 72.23 . Find the length of the longer diagonal. See Fig. 11.22.
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