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Examoogle • User • info@examoogle.com • EE-2024-8821
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From the top of a tower, the angles of depression of two objects, on the ground on the same side of it are observed to be 60° and 30°, respectively, and the height of the tower is 450 m. The distance between the objects (in m) is:
4503
3003
1503
1003
4503
For a tower of height h and angles 60° and 30°, the distance between objects is h × (3 - 1/3), which simplifies to (2h/3) or (2h3)/3.
Tower height h = 450 m, Angle of depression 1 = 60°, Angle of depression 2 = 30°.
d=h(cotθ2−cotθ1)
For a tower of height h and angles 60° and 30°, the distance between objects is h × (3 - 1/3), which simplifies to (2h/3) or (2h3)/3.
Many students confuse angle of elevation and depression, but since they are complementary, they result in the same triangle geometry; do not mistake the 30° angle for the one closer to the tower.
Define Geometric Components
Let height of tower AB = 450 m. Let the two objects be at C (60°) and D (30°). In △ABC, tan60°=AB/BC. In △ABD, tan30°=AB/BD.
BC=3450=1503,BD=4503
Calculate Distance between objects
The distance between the two objects is CD=BD−BC.
CD=4503−1503=3003
B is correct because the calculated distance between the two objects is 3003 m.
This concept of distance between two points based on tower height is a staple in civil engineering surveying and RPF/ALP technical exam sections.