Imagine a climber clipping onto the rope described in Example 7 and pulling himself to the rope’s midpoint. Because the rope is supporting the weight of the climber, it no longer takes the shape of the catenary y = 200 cosh(x/200). Instead, the rope (nearly) forms two sides of an isosceles triangle. Compute the sag angle θ illustrated in the figure, assuming that the rope does not stretch when weighted. Recall from Example 7 that the length of the rope is 101 ft.

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ул Sag angle 0 100+ - 50 х 50