A firefighter on the ground sees fire break through a window near the top of the building. There is voice contact between the ground and firefighters on the roof. The angle of elevation to the windowsill is 28o. The angle of elevation to the roof is 42o. The firefighter is 75 feet from the building and her eyes are 5 feet above the ground. What roof to windowsill distance can she report to the firefighters on the roof?
Answers
Answer:
The difference between the heights is the roof-to-windowsill
Answer:
The answer is 34.1 feet.
Step-by-step explanation:
Let's assume that the distance between the firefighter on the ground and the roof is "x" feet and the distance between the roof and the windowsill is "y" feet. Then, we can use trigonometry to solve for "y".
From the given information, we know that:
The firefighter standing on the ground is 28 degrees above the windowsill.
The angle of elevation to the roof from the firefighter on the ground is 42 degrees.
The distance between the firefighter on the ground and the building is 75 feet.
The height of the firefighter's eyes above the ground is 5 feet.
Using trigonometry, we can write:
tan(28) = (y + 5) / 75 (1) [using the angle of elevation to the windowsill]
tan(42) = (y + 5 + x) / 75 (2) [using the angle of elevation to the roof]
We want to solve for "y", so let's rearrange equation (1) to get:
y = 75 tan(28) - 5
When we enter this expression in place of "y" in equation (2), we obtain:
tan(42) = (75 tan(28) - 5 + x) / 75
Simplifying this equation, we get:
x = 75(tan(42) - tan(28)) + 5
By evaluating this expression, we can determine:
x ≈ 62.7 feet
Therefore, the roof to windowsill distance that the firefighter on the ground can report to the firefighters on the roof is:
y = x tan(28) ≈ 34.1 feet
Hence, the answer is 34.1 feet (rounded to one decimal place).
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