The angle of elevation to the top of the hotel is 55°. If the height of the hotel is 90 feet, about how far away is Mr. Jameson from the hotel?

Question

Reginald Ryan

Mathematics

28 September, 00:44

The angle of elevation to the top of the hotel is 55°. If the height of the hotel is 90 feet, about how far away is Mr. Jameson from the hotel?

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Carley Lloyd · Answer 1

To solve this, use SOH CAH TOA. Your angle would be at Jameson's location, meaning that the hotel's height would be your opposite side and the distance from the hotel would be your adjacent side. This gives you tan (x) = (OPP) / (ADJ). Now, isolate the adjacent part of the equation, ADJ*tan (x) = OPP, ADJ = (OPP) / (tan (x)), plug in your values, ADJ=90 / (tan (55)) ≈90/1.428≈63.019. This means jameson is roughly 63.019 feet away from the hotel.