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Unformatted text preview: arellano (aa39398) – Homework 3 – Weathers – (17101) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Two sides of a triangle are 12 . 5 cm and 31 . 1 cm, with the included angle 105 ◦ . What is the length of the third side? Correct answer: 36 . 3963 cm. Explanation: Let : a = 12 . 5 cm and b = 31 . 1 cm . a c b A C We are looking for the side opposite the given angle so, applying the law of Cosines c 2 = a 2 + b 2 2 a b cos C = (12 . 5 cm) 2 + (31 . 1 cm) 2 2 (12 . 5 cm) (31 . 1 cm) cos 105 ◦ = 1324 . 69 c = √ 1324 . 69 = 36 . 3963 cm . 002 (part 2 of 2) 10.0 points What is angle opposite the side 12 . 5 cm? Correct answer: 19 . 3743 ◦ . Explanation: sin A a = sin C c sin A = a sin C c A = arcsin parenleftbigg a sin C c parenrightbigg = arcsin bracketleftbigg (12 . 5 cm) sin 105 ◦ 36 . 3963 cm bracketrightbigg = 19 . 3743 ◦ . 003 10.0 points A high fountain of water is located at the center of a circular pool as in the figure. A student walks around the pool and estimates its circumference to be 97 . 7 m. Next, the student stands at the edge of the pool and uses a protractor to gauge the angle of elevation of the top of the fountain to be 32 ◦ ....
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This note was uploaded on 10/31/2010 for the course PHYSICS 1710 taught by Professor Weathers during the Fall '10 term at North Texas.
 Fall '10
 Weathers
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