This Article consist of most important ” Height and Distance Questions and Answers ” that are mostly asked in all competitive exams. We collected these questions from the students who appeared in exams. Now try to solve these questions.
Height and Distance Questions and Answers
Question : 1 An observer 2 m tall is 10√3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
A. none of these
B. 12 m
C. 14 m
D. 10 m
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Correct Answer : C
Explanation :
SR = PQ = 2 m
PS = QR = 103√ mtan 30°=TSPS13√=TS103√TS=103√3√=10 m
TR = TS + SR = 10 + 2 = 12 m
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Question : 2 . The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
A. none of these
B. 60°
C. 45°
D. 30°
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Correct Answer : C
Explanation :
Consider the diagram shown above where QR represents the tree and PQ represents its shadow
We have, QR = PQ
Let angleQPR = θ
tan θ = QRPQ=1 (since QR = PQ)
=> θ = 45°
i.e., required angle of elevation = 45°
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Question : 3 A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 45º. What is the distance between the base of the tower and the point P?
A. 9 units
B. 33√ units
C. Data inadequate
D. 12 units
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Correct Answer : C
Explanation :
two equations and 3 variables. Hence we can not find the required value with the given data.
(Note that if one of SR, PQ, QR is known, this becomes two equations and two variables and if that was the case, we could have found out the required value)
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Question : 4 Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 300 m
B. 173 m
C. 273 m
D. 200 m
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Correct Answer : C
Explanation :
Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m, angle BAD = 30° , angle BCD = 45°
tan 30° = BDBA⇒13√=100BA⇒BA=1003√
tan 45° = BDBC⇒1=100BC⇒BC=100
Distance between the two ships = AC = BA + BC
=1003√+100=100(3√+1)=100(1.73+1)=100×2.73=273 m
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