Subject
Quantitative Aptitude
Topic
Speed & Distance
An observer 6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:
An observer 6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:
ক)
21.6 mখ)
23.2 mগ)
24.72 mঘ)
None of theseExplanation
Let AB be the observer and CD be the tower. Draw BE^CD. Then, CE = AB = 1.6 m, BE = AC = 20√3 m. DE/ BE = tan 30º = 1/√3 => DE = 20√3/√3m = 20 m. So, CD = CE+DE = (1.6+20) m = 21.6 m.Related questions
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