Points A and B are 100 m apart andare of the same elevation as the footof a building. The angles of elevationof the top of the building from pointsA and B are 21° and 32° respectively.How far is A from the building inmeters?
Question
Points A and B are 100 m apart andare of the same elevation as the footof a building. The angles of elevationof the top of the building from pointsA and B are 21° and 32° respectively.How far is A from the building inmeters?
Solution
To solve this problem, we can use the tangent of the angle of elevation, which is the ratio of the opposite side (height of the building) to the adjacent side (distance from the point to the building).
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Let's denote the height of the building as h, the distance from point A to the building as x, and the distance from point B to the building as (100 - x).
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From the problem, we know that tan(21°) = h / x and tan(32°) = h / (100 - x).
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We can solve these two equations simultaneously to find the value of x.
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First, express h in terms of x from the first equation: h = x * tan(21°).
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Substitute h in the second equation: x * tan(21°) = (100 - x) * tan(32°).
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Solve this equation for x: x = 100 * tan(32°) / (tan(21°) + tan(32°)).
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Using a calculator, we find that x ≈ 57.7 m.
So, point A is approximately 57.7 meters away from the building.
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