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Height and Distance problems
1. An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The heights of the tower is:
Options
A. 21.6 m
B. 23.2 m
C. 24.72 m
D. None of these
Also asked in:
AIEEE
,
Bank Exams
,
CAT
,
Analyst
,
Bank Clerk
,
Bank PO
Show Answer
Scratch Pad
Discuss
Correct Answer: 21.6 m
Explanation:
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
= tan 30° =
1
BE
√3
⟹ DE =
20√3
m = 20 m.
√3
∴ CD = CE + DE = (1.6 + 20) m = 21.6 m.
2. An observer 1.6 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:
Options
A. 21.6 m
B. 23.2 m
C. 24.72 m
D. None of these
Also asked in:
AIEEE
,
Bank Exams
,
CAT
,
Analyst
,
Bank Clerk
,
Bank PO
Show Answer
Scratch Pad
Discuss
Correct Answer: 21.6 m
First
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