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Quiz 9 Math 1151, Precalculus 2 Name: Summer 2012, Section 002 Problem 1a.(5 points) What is the law of cosines? c2 = a2 + b2 − 2ab cos(C) Problem 1b.(5 points) What is the law of sines? sin A a = sin B b = sin C c Problem 2.(10 points) Solve the triangle which has sides of lengths a=9, b=7, c=10. (Round to the nearest tenth) This is a SSS triangle so we must use the law of cosines to start solving it. 2 −a2 −b2 C = cos−1 ( c −2ab 2 2 2 −7 −9 ) = cos−1 ( 10−2(9)(7) ) = cos−1 (5/21) = 76.2◦ . 2 −c2 −b2 We can continue to use the law of cosines to find one more angle: A = cos−1 ( a 2 2 2 −10 −9 cos−1 ( 9−2(10)(7) ) = cos−1 (17/35) = 60.9◦ . And lastly we find B = 180◦ − 60.9◦ − 76.2◦ = 42.9◦ . −2cb )= Problem 3.(10 points) An airplane is spotted by two people, one at point P and the other at point Q, who are 1000 feet apart. At one instance each person takes a sighting of the angle of elevation to the plane as indicated in the illustration. How high is the plane (find h)? This is currently a ASA triangle. So we are going to use the law of sines to solve it. We first start by solving for the last angle which is 180◦ − 85◦ − 70◦ = 25◦ . Now we can use the law of sines to solve for a or b: b = 1000 sin(70◦ ) sin(25◦ ) 1000 sin(85◦ ) sin(25◦ ) = 2357.2 ft and a = = 2223.5 ft. To find the height of the plane we will use one of the right triangles: h = b sin(70◦ ) = a sin(85◦ ) = 2215.04 ft. Extra Credit(10 points) Find a formula for all the vertical asymptotes of cot(2x). Then list all the vertical asymptotes (in radians) that lie in the interval (−2π, 0). In order to find the asymptotes we need to solve for when the denominator of cot(2x) is zero. So we need to solve the trigonometric equation sin(2x) = 0. This occurs when 2x = πk. Solving for x we get an equation that tell us where all the vertical asymptotes: x = π2 k where k is an integer.