auto generate barcode vb net Fig. 11.1 in VS .NET

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Fig. 11.1
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11.2 Law of Sines
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In any triangle ABC, the ratio of a side and the sine of the opposite angle is a constant; i.e., a sin A b sin B c sin C or sin A a sin B b sin C c
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For a proof of the law of sines, see Prob. 11.1.
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11.3 Law of Cosines
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In any triangle ABC, the square of any side is equal to the sum of the squares of the other two sides diminished by twice the product of these sides and the cosine of the included angle; i.e., a2 b2 c2 b2 a2 a2 c2 c2 b2 2bc cos A 2ac cos B 2ab cos C
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For a proof of the law of cosines, see Prob. 11.3.
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CHAPTER 11 Oblique Triangles
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11.4 Solution of Oblique Triangles
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When three parts of a triangle, not all angles, are known, the triangle is uniquely determined, except in one case noted below. The five cases of oblique triangles are
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Case I: Given two angles and the side opposite one of them Case II: Given two angles and the included side Case III: Given two sides and the angle opposite one of them Case IV: Given two sides and the included angle Case V: Given the three sides
Case I II III IV V Use Law of Sines Sines Sines Cosines Cosines First Part to Find Side opposite second given angle Third angle, then either of remaining sides Angle opposite second given side Third side Any angle can be found
In Case III there is not always a unique solution. It is possible to have no solution for the angle, one solution for the angle, or two solutions an angle and its supplement. See Example 11.3 and Prob. 11.2 for a complete discussion of this case.
Case I
Given two angles and the side opposite one of them
EXAMPLE 11.1 Suppose b, B, and C are given.
To find c, use
c b ; then c sin C sin B To find A, use A 180 (B C). To find a, use a sin A b ; then a sin B
b sin C . sin B b sin A . sin B
(See Prob. 11.4.)
Case II
Given two angles and the included side
EXAMPLE 11.2 Suppose a, B, and C are given.
To find A, use A To find b, use
C). a sin B . sin A a sin C . sin A
b sin B c To find c, use sin C
a ; then b sin A a ; then c sin A
(See Prob. 11.5.)
Case III
Given two sides and the angle opposite one of them
CHAPTER 11 Oblique Triangles
EXAMPLE 11.3 Suppose b, c, and B are given.
sin B c sin B , sin C . b b If sin C > 1, no angle C is determined. If sin C 1, C 90 and a right triangle is determined. If sin C < 1, two angles are determined: an acute angle C and an obtuse angle C 180 one or two triangles determined. If C B 180 , then the angle C is not a solution. From sin C c
C. Thus, there may be
This case is discussed geometrically in Prob. 11.2. The results obtained may be summarized as follows: When the given angle is acute, there will be (a) One solution if the side opposite the given angle is equal to or greater than the other given side (b) No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side When the given angle is obtuse, there will be (c) No solution when the side opposite the given angle is less than or equal to the other given side (d) One solution if the side opposite the given angle is greater than the other given side
EXAMPLE 11.4
(1) When b
(2) When b
30, c 20, c
20, and B 30, and B
40 , there is one solution since B is acute and b > c. 40 , there is either no solution, one solution, or two solutions. The particular subc sin B . b 140 , there is one solution. 140 , there is no solution.
case is determined after computing sin C (3) When b (4) When b 30, c 20, c 20, and B 30, and B
This, the so-called ambiguous case, is solved by the law of sines. (See Probs. 11.9 to 11.11.)
Case IV
Given two sides and the included angle
EXAMPLE 11.5 Suppose a, b, and C are given.
To find c, use c2
To find A, use sin A To check, use A B
b2 2ab cos C. a sin C c . To find B, use sin B C 180 .
b sin C c .
(See Probs. 11.13 and 11.14.)
Case V
Given the three sides
EXAMPLE 11.6 With a, b, and c given, solve the law of cosines for each of the angles.
To find the angles, use cos A To check, use A B C
b2 180 .
c2 2bc
, cos B
a2 2ca
, and cos C
b2 2ab
(See Probs. 11.17 and 11.18.)
CHAPTER 11 Oblique Triangles
SOLVED PROBLEMS
11.1 Derive the law of sines.
Fig. 11.2
Fig. 11.3
Let ABC be any oblique triangle. In Fig. 11.2, angles A and B are acute, while in Fig. 11.3, angle B is obtuse. Draw CD perpendicular to AB or AB extended and denote its length by h. In the right triangle ACD of either figure, h b sin A, while in the right triangle BCD, h a sin B since in Fig. 11.3, h a sin / DBC a sin (180 B) a sin B. Thus, a sin B b sin A or a sin A b sin B
In a similar manner (by drawing a perpendicular from B to AC or a perpendicular from A to BC), we obtain a sin A Thus, finally, a sin A b sin B c sin C c sin C b sin B c sin C
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