From his eye, which stands 1.63 meters above the ground, Isaac measures the angle of elevation to the top of a prominent skyscraper to be 17degrees ∘ . If he is standing at a horizontal distance of 294 meters from the base of the skyscraper, what is the height of the skyscraper? Round your answer to the nearest hundredth of a meter if necessary.
Question
From his eye, which stands 1.63 meters above the ground, Isaac measures the angle of elevation to the top of a prominent skyscraper to be 17degrees ∘ . If he is standing at a horizontal distance of 294 meters from the base of the skyscraper, what is the height of the skyscraper? Round your answer to the nearest hundredth of a meter if necessary.
Solution
To solve this problem, we can use the tangent of the angle of elevation, which is the ratio of the opposite side (the height of the skyscraper) to the adjacent side (the distance from Isaac to the skyscraper).
The formula for the tangent of an angle in a right triangle is:
tan(θ) = opposite/adjacent
We know that the angle of elevation (θ) is 17 degrees and the distance from Isaac to the skyscraper (adjacent) is 294 meters. We want to find the height of the skyscraper (opposite), which we'll call h.
So we can set up the equation:
tan(17) = h/294
To solve for h, we multiply both sides by 294:
h = 294 * tan(17)
Using a calculator, we find that tan(17) is approximately 0.30573, so:
h = 294 * 0.30573 ≈ 89.88 meters
However, this is the height from Isaac's eye level to the top of the skyscraper. Since Isaac's eye level is 1.63 meters above the ground, we need to add this to our calculated height to get the total height of the skyscraper:
Total height = h + Isaac's height = 89.88 m + 1.63 m = 91.51 m
So, the height of the skyscraper is approximately 91.51 meters, rounded to the nearest hundredth of a meter.
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