Thus, that hypotenuse has length 1000 feet. This leads to thevery simple equation x = 1000 sin(60°). In either case, you willget x≈866.03 feet and so h≈871.03 feet.[Warning: The cuteisosceles triangle thingy doesn’t usually happen. The otherapproach 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 thepoint. In doing this, make clear which values are exact andwhich 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 aregiven by (x,y) = (-5,-5√3), obtain polar coordinates for thepoint.It’s easy to see r2= 100. Thus, use r = 10, to keep thingssimple. It turns out that the reference angleθrsatisfies theequation tan(θr) = 31/2. Thus, because the point lies in thethird quadrant, we may use eitherθ= 240° orθ= 4π/3. Fromhere, it is easy to list all pairs (r,θ) that represent thepoint.16. (10 pts.) (a) Obtain all solutions to the equation below,and then (b) list the solutionsθwith 0≤θ< 2π.2 sin2(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. Allsolutions to sin(θ) = -1/2 are given byθ= 7π/6 + 2kπorθ= 11π/6 + 2kπ, k any integer, and all solutions to sin(θ) = -1are given byθ= 3π/2 + 2kπ, k any integer. (b) The solutions inthe desired interval are 7π/6, 3π/2, and 11π/6.
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