auto generate barcode vb net Discuss the several special cases when two sides and the angle opposite one of them are given. in .NET framework

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11.2 Discuss the several special cases when two sides and the angle opposite one of them are given.
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Let b, c, and B be the given parts. Construct the given angle B and lay off the side BA c. With A as center and radius equal to b (the side opposite the given angle), describe an arc. Fig. 11.4(a) to (e) illustrates the special cases which may occur when the given angle B is acute, while Fig. 11.4( ) and (g) illustrates the cases when B is obtuse. The given angle B is acute. Fig. 11.4(a). When b < AD c sin B, the arc does not meet BX and no triangle is determined. Fig. 11.4(b). When b AD, the arc is tangent to BX and one triangle a right triangle with the right angle at C is determined. Fig. 11.4(c). When b > AD and b < c, the arc meets BX in two points C and C on the same side of B. Two triangles ABC, in which C is acute, and ABC , in which C 180 C is obtuse are determined. Fig. 11.4(d). When b > AD and b c, the arc meets BX in C and B. One triangle (isosceles) is determined. Fig. 11.4(e). When b > c, the arc meets BX in C and BX extended in C . Since the triangle ABC does not contain the given angle B, only one triangle ABC is determined. The given angle is obtuse. Fig. 11.4( ). When b < c or b c, no triangle is formed. Fig. 11.4(g). When b > c, only one triangle is formed as in Fig. 11.4(e).
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CHAPTER 11 Oblique Triangles
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Fig. 11.4
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11.3 Derive the law of cosines.
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In either right triangle ACD of Fig. 11.5, b2 In the right triangle BCD of Fig. 11.5(a), h Then and AD b2 AB h2
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(AD)2. a cos B. a2 cos2 B
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a sin B and DB c2
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a cos B a2 sin2 B 2ca cos B c c 2ca cos B a2 2ca cos B
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a (sin B
cos B)
Fig. 11.5
In the right triangle BCD of Fig. 11.5(b), h and Then BD AD a sin / CBD a cos / CBD AB BD c a sin (180 a cos (180 a cos B B) B) and a sin B a cos B b2 c2 a2 2ca cos B
The remaining equations may be obtained by cyclic changes of the letters.
CHAPTER 11 Oblique Triangles
Case I 11.4 Solve the triangle ABC, given a 62.5, A 112 20 , and C 42 10 . See Fig. 11.6.
Fig. 11.6
For B:
180 a sin B sin A
For b: b
154 30 25 30 62.5(0.4305) 62.5 sin 25 30r 29.1 0.9250 sin 112 20r 112 20 ) sin 67 40 ] 62.5(0.6713) 0.9250 45.4
[sin 112 20 For c: c a sin C sin A
sin (180
62.5 sin 42 10r sin 112 20r 29.1, c
The required parts are b
45.4, and B
25 30 .
Case II 11.5 Solve the triangle ABC, given c 25, A 35 , and B 68 . See Fig. 11.7.
Fig. 11.7
For C: For a: For b:
C a b
180 c sin A sin C c sin B sin C
B) 25 sin 35 sin 77 25 sin 68 sin 77 15, b
103 77 25(0.5736) 15 0.9744 25(0.9272) 0.9744 24
The required parts are a
24, and C
11.6 A and B are two points on opposite banks of a river. From A, a line AC 275 m is laid off, and the angles CAB 125 40 and ACB 48 50 are measured. Find the length of AB.
In the triangle ABC of Fig. 11.8(a), B AB c b sin C sin B 180 (C A) 5 30 and 275(0.7528) 0.0958 2160 m
275 sin 48 50r sin 5 30r
CHAPTER 11 Oblique Triangles
Fig. 11.8
11.7 A ship is sailing due east when a light is observed bearing N62 10 E. After the ship has traveled 2250 m, the light bears N48 25 E. If the course is continued, how close will the ship approach the light (See Prob. 5.5, Chap. 5.)
Refer to Fig. 11.8(b). In the oblique triangle ABL: AB / ALB BL In the right triangle BLC: BL CL 2250, / BAL 180 (/ BAL AB sin / BAL sin /ALB 4420 and / CBL BL sin / CBL 27 50 , and / ABL / ABL) 13 45 2250(0.4669) 0.2377 41 35 4420(0.6637) 2934 m 4420 138 25
2250 sin 27 50r sin 13 45r 90 48 25
4420 sin 41 35
For an alternative solution, find AL in the oblique triangle ABL and then CL in the right triangle ALC.
11.8 A tower 125 ft high is on a cliff on the bank of a river. From the top of the tower, the angle of depression of a point on the opposite shore is 28 40 , and from the base of the tower, the angle of depression of the same point is 18 20 . Find the width of the river and the height of the cliff.
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