Aptitude - Height and Distance
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- Height and Distance - Formulas
- Height and Distance - General Questions
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. |
One of AB, AD and CD must have given.
So, the data is inadequate.
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. |
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.
Let AB be the tower.
Then, APB = 30° and AB = 100 m.
AB | = tan 30° = | 1 |
AP | 3 |
![]() |
= (AB x 3) m |
= 1003 m | |
= (100 x 1.73) m | |
= 173 m. |