Discussion :: Height and Distance
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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 30°. The height of the tower is:
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.
\(\frac { DE } { BE } \)= tan 30° = \(\frac {1} { 3 } \) | ||
DE = | \(\frac { 203 } { 3 } \)m = 20 m. | |
CD = CE + DE = (1.6 + 20) m = 21.6 m.
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