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Solve the Problem dft\mathrm { d } \mathrm { ft }

Question 116

Multiple Choice

Solve the problem.
-It can be shown that if the angle of elevation from an observer to the top of an object is A and the angle of elevation dft\mathrm { d } \mathrm { ft } closer is B\mathrm { B } , then the distance from the object to the closest point of observation is given by
D=dcot⁡Bcot⁡A−cot⁡Bft. \mathrm { D } = \frac { \mathrm { d } \cot \mathrm { B } } { \cot \mathrm { A } - \cot \mathrm { B } } \mathrm { ft } \text {. }
Find BB if D=40ft,d=20ftD = 40 \mathrm { ft } , \mathrm { d } = 20 \mathrm { ft } , and A=33∘A = 33 ^ { \circ } . Give your answer in degrees to the nearest hundredth.


A) 44.25∘44.25 ^ { \circ }
B) 33.00∘33.00 ^ { \circ }
C) 62.83∘62.83 ^ { \circ }
D) 17.99∘17.99 ^ { \circ }

Correct Answer:

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