154E3-S2010

154E3-S2010 - MA 15400 Spring 2010 Exam 3 C LAW OF SINES...

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MA 15400 Spring 2010 Exam 3 LAW OF SINES LAW OF COSINES sin α a = sin β b = sin γ c c 2 = a 2 + b 2 2 ab cos ANGLE BETWEEN TWO VECTORS: cos θ = ( r a ) ( r b ) r a r b α β γ A B C c a b
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MA 15400 Exam 3 Spring 2010 Covers Lessons 21 to 34, all of Section 7.6, 8.1, 8.2, 8.3, 8.4, and 4.5 1 1. Find the exact value of the expression. 1 2 sin sin 3 π A. 3 B. 2 3 C. 3 D. 4 3 E. None of the above 2. Approximate the solutions of the equation, to four decimals, in the interval , 2 2 π π . 2 5sin sin 2 0 x x = A. 0.7373, 0.9633 B. 0.5708, 0.8335 C. 0.4820, 1.4245 D. 0.5403, 0.7403 E. None of the above
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MA 15400 Exam 3 Spring 2010 Covers Lessons 21 to 34, all of Section 7.6, 8.1, 8.2, 8.3, 8.4, and 4.5 2 3. Find the exact value of the expression. 1 cos 2sin 5 x A. 2 10 2 25 x x B. 5 2 5 x C. 2 2 25 25 x x D. 2 25 2 25 x E. None of the above 4. There are two distinct triangles possible with a side a = 12.0 cm , side b = 17.0 cm , and angle 38 α = ° . Find the perimeter of both triangles to the nearest tenth of a
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154E3-S2010 - MA 15400 Spring 2010 Exam 3 C LAW OF SINES...

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