Question

Kayson is looking at two buildings, building A and building B, at an angle of elevation of 40°. Building A is 120 feet away, and building B is 60 feet away. Which building is taller and by approximately how many feet?
Building A; it is around 100.69 feet taller than building B
Building A; it is around 50.34 feet taller than building B
Building B; it is around 100.69 feet taller than building A
Building B; it is around 50.34 feet taller than building A

Answer 1

The right angle from Kayson‘s perspective looking at each building establishes the relationship between the height of the building (opposite to Kayson)/distance to the building (adjacent to Kayson). The ratio of the opposite versus adjacent sides of that right angle is called the tangent. The tangent of 40° equals 0.839. Building A is 120 feet away (adjacent to Kayson), so the height of building A (opposite to Kayson) is 0.839×120 = 100.68. Likewise, The height of building B would be 0.839×60 = 50.34. Therefore, building A is 100.68-50.34=50.34‘ taller than building B.

Answer 2

Answer: Choice B

Building A; it is around 50.34 feet taller than building B

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Explanation:

Draw two right triangles such that they both have a reference angle of 40 degrees. Since it's an angle of elevation, place the angle near the horizontal side.

Opposite each reference angle is the height of the building, which we'll call h.

The adjacent side is the horizontal distance from the person to the building.

Building A is 120 ft away so it's height is...

tan(angle) = opposite/adjacent

tan(40) = h/120

h = 120*tan(40)

h = 100.69

This value is approximate. Make sure your calculator is in degree mode.

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Repeat the same idea for building B, except the adjacent side is smaller this time.

tan(angle) = opposite/adjacent

tan(40) = h/60

h = 60*tan(40)

h = 50.35

This value is approximate.

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Now subtract the heights

100.69 - 50.35 = 50.34

This shows that building A is roughly 50.34 taller than building B.