A firefighter on the ground sees a fire break through a window near the top of a building. The angle of elevation to the windowsill is 28°. The angle of elevation to the top of the building is 42°. The firefighter is 75 ft. from the building and her eyes are 5 ft. above the ground.
a. about 28 ft.
b. about 33 ft.
c. about 40 ft.
d. about 68 ft.
In this problem, there are two right triangles formed. One is formed with the height of the building as its vertical side, and the other is formed with the height of the windowsill as its vertical side. We can use the tangent function to solve both triangles from the angle of elevation information.
Let's call the height of the building, H and the height to the windowsill, W.
For the building height,
tan(42°) = H / 75
H = 75 * tan(42°)
For the windowsill height,
tan(28°) = W / 75
W = 75 * tan(28°)
Now, we want to find the difference between these two heights (ΔH) and also account for the firefighter's eye level which is 5 ft. above the ground.
ΔH = H - W
ΔH = (75 * tan(42°)) - (75 * tan(28°))
Calculating the values and subtracting,
ΔH = (75 * 0.9004) - (75 * 0.5317)
ΔH = 67.53 - 39.8775
ΔH = 27.6525
However, we need to include the height from the ground to firefighter's eyes, so we add that 5 ft. to the difference in heights:
ΔH = 27.6525 + 5
ΔH = 32.6525
Hence, the height of the windowsill above the firefighter's eyes is about 33 ft. The correct answer is choice b.