MAC
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# 11 10 pts use the law of sines to solve the triangle

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11. (10 pts.) Use the Law of Sines to solve the triangle with α = 110°, γ = 30°, and c = 6. You may assume that the standard labelling scheme is used. The missing pieces are β = 40°, a = 12sin(110°) 11.28, and b = 12sin(40°) 7.71. 12. (5 pts.) Determine whether one, two, or no triangles result from the following data. You do not have to solve the triangles that might result. You may assume that the standard labelling scheme is used. a = 3, b = 6, α = 32° If there were such a triangle, we would have to have sin(32°)/3 = sin( β )/6 true for some β . Unfortunately, this last equation is equivalent to sin( β ) = 2 sin(32°) > 2 sin(30°) = 1. There is no solution to this, and thus, no triangle.

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TEST-03/MAC1114 Page 4 of 4 13. (10 pts.) To measure the height of the top of a distant object on a level plane, a surveyor takes two sightings of the top of the object 1000 feet apart. The first sighting, which is nearest the object, results in an angle of elevation of 60°. The second sighting, which is most distant from the object, results in an angle of elevation of 30°. If the transit used to make the sightings is 5 feet tall, what is the height of the object. [Hint: Make a diagram of the situation. The distance from the base of the object is unknown. ] Let h denote the height of the object, x the length of the side opposite the 60° sighting, and d the length of the side adjacent to the 60° angle. The sides of length d and x meet in a right angle. Then we have h = x + 5, and the system of equations tan(60°) = x/d and tan(30°) = x/(d + 1000). Solving for x yields x = 1000(tan(30°))/(1 - cot(60°)tan(30°)). Alternatively, you could recognize that, here, the hypotenuse of the triangle created by the 60° sighting is one of the two equal sides of the isosceles 30°,30°,120° triangle formed from the 30° sighting.
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