Question

amari is standing 50 feet from the base of a building. from where he stands, the angle formed between the top of the building and the ground at his feet is 60°. how tall is the building? 50 ft, $\frac{50sqrt{3}}{3}$ ft, $50sqrt{3}$ ft, 100 ft

Explanation:

Step1: Use tangent function

\(\tan\theta=\frac{opposite}{adjacent}\), here \(\theta = 60^{\circ}\), adjacent = 50 ft. Let height above 50 - ft level be \(h\). So \(\tan60^{\circ}=\frac{h}{50}\).

Step2: Solve for \(h\)

Since \(\tan60^{\circ}=\sqrt{3}\), then \(h = 50\sqrt{3}\) ft. Total height of building is \(50 + 50\sqrt{3}\) ft. But if we assume the 50 - ft is part of the right - triangle height calculation, using \(\tan60^{\circ}=\frac{h}{50}\), \(h = 50\sqrt{3}\) ft.

Answer:

\(50\sqrt{3}\text{ ft}\)