Multiple choice

Two buildings AB and CD are such that height AB is greater than height CD. The perpendicular distance between two buildings is 100 m. The angle of elevation of top of building AB from the top of building CD is 60o. The distance between tops of buildings will be

  1. 100 m

  2. 200 m

  3. 300 m

  4. 400 m

  5. 500 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the height difference be h. tan(60) = h / 100, so h = 100 * sqrt(3). The distance between the tops is the hypotenuse of a triangle with base 100 and height h. Distance = sqrt(100^2 + (100*sqrt(3))^2) = sqrt(10000 + 30000) = sqrt(40000) = 200.

AI explanation

Let the perpendicular distance between the buildings be the adjacent side of a right triangle formed by their tops. Using trigonometric ratios, we apply the formula cos 60 degrees equals the adjacent side divided by the hypotenuse. Substituting the values gives 0.5 equals 100 m divided by the distance between the tops, so the distance between the tops is 200 m.

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