Question

At 11:30 a.m. the angle of elevation of the sun for one city is 55.9 ∘

. If the height of a monument is approximately 558ft, what is the length of the shadow it will cast at that time? Round to the nearest foot. The length of the shadow cast by the monument is approximately ft. An aerial tram takes passengers from an elevation of 6316ft to an elevation of 10,455ft up the slope of a mountain. If the tram runs on a cable measuring 14,863ft in length, what is the angle of incline of the tram to the nearest tenth of a degree? The angle of incline of the tram is approximately

Answer 1

Drawing diagrams

a) tan 55.9 = H / L = 558 / L

L = 378 ft length of shadow of monument

b) H = 10455 - 6316 = 4139 increase in elevation

sin θ = H / L = 4139 / 14863 = .278

θ = 16.2 deg

Answer 2

The length of the shadow cast by the monument at 11:30 a.m. is approximately 900 ft. The angle of incline of the tram is approximately 16.8 degrees.

The length of the shadow cast by the monument can be found using the tangent function. Tan(angle) = Opposite/Adjacent. We know that the angle of elevation of the sun is 55.9 degrees and the height of the monument is 558 ft. Assuming the length of the shadow is 'x', we can set up the equation as Tan(55.9) = x/558. Solving for x, we get x ≈ 900 ft. So, the length of the shadow cast by the monument at 11:30 a.m. is approximately 900 ft.

The angle of incline of the tram can be found using the sine function. Sin(angle) = Opposite/Hypotenuse. We know that the length of the cable is 14,863 ft and the difference in elevation is 10,455 ft - 6,316 ft = 4,139 ft. Assuming the angle of incline is 'A', we can set up the equation as Sin(A) = 4,139/14,863. Solving for A, we get A ≈ 16.8 degrees. So, the angle of incline of the tram is approximately 16.8 degrees.