Aptitude - Height and Distance - Discussion
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Read more:"Everyone is wise until he speaks."
- (Proverb)
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An observer 1.6 m tall is 203 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:
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21.6 m | [B]. |
23.2 m | | [C]. |
24.72 m | [D]. |
None of these |
Answer: Option A
Explanation:
Let AB be the observer and CD be the tower.
Draw BE  CD.
Then, CE = AB = 1.6 m,
BE = AC = 203 m.
 DE = |
203 |
m = 20 m. |
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 CD = CE + DE = (1.6 + 20) m = 21.6 m.
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Yazhini.M.R said:
(Wed, Aug 3, 2011 11:27:23 AM)
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Since tan thetq= opposite/adjacent , tan 30= DE/BE ,
the value of tan 30 is 1/ root 3,
then DE/BE=1/root 3, then DE=BE/root 3,
thus 20 m is obtained.
since CD= CE+DE, 21.6 is obtained. |
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