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 Sines 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: Consider a right triangular wall panel with sides a, b and c , opposite angles A , B and C respectively. Sketch the graph of the triangle with known measures as shown below in Figure 1. From Figure 1, it is observed that the acute angles are A and B . Let area of the right triangular wall panel be denoted as A t . Then, A t = 1 2 a b Use the law of sines and obtain the area as follows

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Chapter 9, Problem 43RE

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To prove: The area of the right triangular wall panel is At=a2sinB2sinA.

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Chapter 9, Problem 42RE

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Chapter 9 Solutions

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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To prove: The area of the right triangular wall panel is A t = a 2 sin B 2 sin A .

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