# 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 = AB x 3 = 1003 m. |

AC | 3 |

AB | = tan 45° = 1 AD = AB = 100 m. |

AD |

CD = (AC + 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 |

BC | = 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 |

DE = | 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 |

AP | = (AB x 3) m |

= 1003 m | |

= (100 x 1.73) m | |

= 173 m. |