A mailbox on the street is 600 feet from the top of a building. From the mailbox looking up to the tops of that building and a second, taller building, the angle between the buildings is 67degrees. The tops of the buildings are 800 feet from each other. If the angle of elevation from the mailbox to the first building is 53 degrees and the taller building is 60 degrees, how tall is each building... Step by step explanation..Please..Thanks

Step-by-step explanation:

Label the mailbox viewpoint, the top of the first building, and the top of the second building points M, A, and B, respectively.

Consider the triangle MAB. You are told length MA is 600 ft, and length AB is 800 ft. Side AB is opposite the 67° angle at M, so we can use the Law of Sines to find other measures of that triangle. Specifically, we want to know the distance MB.

  sin(B)/MA = sin(M)/AB

  sin(B) = (MA/AB)sin(M) = 600/800·sin(67°) ≈ 0.690379

  B ≈ 43.66°

Then the angle at A is ...

  180° -67° -43.66° = 69.34°

and the side MB can be found from ...

  MB/sin(A) = AB/sin(M)

  MB = (sin(A)/sin(M))AB = sin(69.34°)/sin(67°)·800 ≈ 813.2 . . . feet

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Now, label points on the ground below A and B as A' and B'. This problem is asking for the heights AA' and BB'. Each of these is opposite the angle of elevation to points A and B, respectively, so can be found using the sine relation:

  sin(elevation to A) = AA'/MA

  AA' = MA·sin(elevation to A) = 600·sin(53°) ≈ 479.2 . . . ft

Similarly, ...

  BB' = MB·sin(elevation to B) = 813.2·sin(60°) ≈ 704.2 . . . feet

The first building is 479.2 ft tall; the second is 704.2 ft tall.

c. 4 and 1/2 pounds is the answer! : )