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Height and Distance
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 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 173 m
B. 200 m
C. 273 m
D. 300 m
Answer: Option C
Explanation:
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, ACB = 30° and
ADB = 45°.
AB | = tan 30° = | 1 | ![]() |
AC | 3 |
AB | = tan 45° = 1 ![]() |
AD |
![]() |
= (1003 + 100) m |
= 100(3 + 1) | |
= (100 x 2.73) m | |
= 273 m. |
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60°. What is the distance between the base of the tower and the point P?
A. 4√3 units
B. 8 units
C. 12 units
D. Data inadequate
E. None of these
Answer: Option D
Explanation:
One of AB, AD and CD must have given.
So, the data is inadequate.
Answer: Option D
Explanation:
Let AB be the wall and BC be the ladder.
Then, ACB = 60° and AC = 4.6 m.
AC | = cos 60° = | 1 |
BC | 2 |
![]() |
= 2 x AC |
= (2 x 4.6) m | |
= 9.2 m. |
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 height of the tower is:
A. 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 | = tan 30° = | 1 |
BE | 3 |
![]() |
203 | m = 20 m. |
3 |
CD = CE + DE = (1.6 + 20) m = 21.6 m.
Answer: Option C
Explanation:
Let AB be the tower.
Then, APB = 30° and AB = 100 m.
AB | = tan 30° = | 1 |
AP | √3 |
![]() |
= (AB x √3 ) m |
= 100√3 m | |
= (100 x 1.73) m | |
= 173 m. |
Answer: Option A
Explanation:
Let AB be the tree and AC be its shadow.
Let ACB =
.
Then, | AC | = | √3 ![]() ![]() |
AB |
= 30°.