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. What roof-to-windowsill distance can she report by radio to firefighters on the roof?

a. about 28 ft.

b. about 33 ft.

c. about 40 ft.

d. about 68 ft.

1 answer

First, we need to find the height of the window and the height of the building by using the tangent function with the given angles of elevation.

tan(28°) = (window height - 5 ft) / 75 ft
window height = 75 ft * tan(28°) + 5 ft ≈ 47.87 ft

tan(42°) = (building height - 5 ft) / 75 ft
building height = 75 ft * tan(42°) + 5 ft ≈ 69.44 ft

Now we can find the height difference between the rooftop and the windowsill:

roof_to_windowsill_height = building height - window height ≈ 69.44 ft - 47.87 ft ≈ 21.57 ft

Now we can use the Pythagorean theorem to find the windowsill-to-roof distance:

distance² = 75 ft² + (roof_to_windowsill_height)²
distance² = 75² + 21.57² ≈ 6364.64
distance ≈ sqrt(6364.64) ≈ 79.78 ft

Now, we have to subtract the distance from the firefighter to the building (75 ft) to find the distance between the windowsill and the rooftop:

windowsill_to_roof_distance = 79.78 ft - 75 ft ≈ 4.78 ft

The nearest answer is:

a. about 28 ft.