Explanation Formula used: For any triangle with sides a, b and c , opposite angles A , B and C respectively, the law of Cosines and law of Sine’s can be defined as follows. Law of cosines: a 2 = b 2 + c 2 − 2 b c cos A This can be also written as follows. b 2 = a 2 + c 2 − 2 a c cos B c 2 = a 2 + b 2 − 2 a b cos C Law of sines: a sin A = b sin B = c sin C Calculation: Given that, the magnitude of b = 10 , c = 30 and B = 20 ° . Sketch the graph of the triangle with known measures as shown below in Figure 1. From Figure 1, it is observed that the two sides and the angle opposite one of the given side of the triangle is known. Use the law of cosines and the side a can be obtained as follows. Substitute 10for b , 30 for c and 20 ° for B in b 2 = a 2 + c 2 − 2 a c cos B . b 2 = a 2 + c 2 − 2 a c cos B b 2 − c 2 + 2 a c cos B = a 2 a 2 = 10 2 − 30 2 + 2 ( a ) ( 30 ) cos 20 ° On further simplifications, a 2 = 100 − 900 + 60 ( a ) cos 20 ° a 2 = − 800 + 60 ( a ) 0

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ISBN: 9780134508290

Author: Evans

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Chapter 9.6, Problem 22E

To determine

The equations to solve the given triangleusing law of cosines and determine the reason for using law of sines as easy way.

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Chapter 9.6, Problem 21E

Chapter 9.6, Problem 23E

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Chapter 9 Solutions

EBK BASIC TECHNICAL MATHEMATICS

Ch. 9.1 - For the vectors in Example 2, show that R = B + C...Ch. 9.1 - Prob. 2PE Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E

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