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.
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.