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?

Question

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?

Answer 1

Let's call the web between the firefighter and the bottom of the building (75 ft. away) A, the top of the building B, and the window C. We want to find the distance between point B and point C. First, we need to find the heights of point C (the window) and point B (the top of the building) above her eyes.

We can use tangent to find these heights. The tangent of an angle in a right triangle is the ratio of the side opposite the angle to the side adjacent to the angle (in this case, the height above her eyes to the distance she is from the building).

1. For the window (height h1):
tan(28°) = h1 / 75
h1 = 75 * tan(28°)

2. For the top of the building (height h2):
tan(42°) = h2 / 75
h2 = 75 * tan(42°)

Now we can subtract h1 from h2 to get the height difference (h) between the top of the building and the window.
h = h2 - h1

And since her eyes are 5 ft. above the ground, we need to add 5 ft. to both h1 and h2.
h1_adj = h1 + 5
h2_adj = h2 + 5

Next, we'll use the Pythagorean Theorem to find the hypotenuse (roof-to-windowsill distance) that we're looking for:

roof_distance^2 = (roof_height - window_height)^2 + (distance_from_building)^2
= (h2_adj - h1_adj)^2 + 75^2

Now we can plug in the values we found earlier and solve for the roof_distance.

roof_distance = sqrt((h2_adj - h1_adj)^2 + 75^2)

Let's calculate the values:

h1 = 75 * tan(28) ≈ 39.65 ft
h2 = 75 * tan(42) ≈ 64.95 ft
h1_adj = h1 + 5 ≈ 44.65 ft
h2_adj = h2 + 5 ≈ 69.95 ft

Now we can find the roof-to-windowsill distance:

roof_distance = sqrt((69.95 - 44.65)^2 + 75^2)
roof_distance = sqrt(25.3^2 + 75^2)
roof_distance ≈ 78.6 ft

The firefighter can report a roof-to-windowsill distance of about 78.6 ft to the firefighters on the roof.