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# Alternatively you could recognize that here the

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e 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. Thus, that hypotenuse has length 1000 feet. This leads to the very simple equation x = 1000 sin(60°). In either case, you will get x 866.03 feet and so h 871.03 feet. [Warning: The cute isosceles triangle thingy doesn’t usually happen. The other approach is more general.] 14. (5 pts.) If the polar coordinates of a point are given by (r, θ ) = (9.5,110°), find the rectangular coordinates for the point. In doing this, make clear which values are exact and which are approximations. (x,y) = ((9.5)cos(110°),(9.5)sin(110°)) (-3.25, 8.93) 15. (5 pts.) If the rectangular coordinates of a point are given by (x,y) = (-5,-5 √ 3), obtain polar coordinates for the point. It’s easy to see r 2 = 100. Thus, use r = 10, to keep things simple. It turns out that the reference angle θ r satisfies the equation tan( θ r )=3 1/2 . Thus, because the point lies in the third quadrant, we may use either θ = 240° or θ =4 π /3. From here, it is easy to list all pairs (r, θ ) that represent the point. 16. (10 pts.) (a) Obtain all solutions to the equation below, and then (b) list the solutions θ with 0 ≤θ <2 π . 2 sin 2 ( θ ) + 3 sin( θ )+1=0 (a) The given equation is equivalent to (2sin( θ ) + 1)(sin( θ )+1 )=0 , which is equivalent to sin( θ ) = -1/2 or sin( θ ) = -1. All solutions to sin( θ ) = -1/2 are given by θ =7 π / 6+2 k π or θ =1 1 π / 6+2 k π , k any integer, and all solutions to sin( θ )=- 1 are given by θ
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