# Height and Distance

## Height and Distance

Congratulations - you have completed *Height and Distance*.

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Your answers are highlighted below.

Question 1 |

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:

173 m | |

200 m | |

273 m | |

300 m |

Question 1 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/AC = tan 30° = 1/squareroot(3) ==> AC = AB x squareroot(3) = 1003 m.
AB/AD = tan 45° = 1 AD = AB = 100 m.
==> CD = (AC + AD) = (100*squareroot(3) + 100) m
= 100(squareroot(3) + 1)
= (100 x 2.73) m
= 273 m.

Question 2 |

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 60º. What is the distance between the base of the tower and the point P?

43 units | |

8 units | |

12 units | |

Data inadequate |

Question 2 Explanation:

the data is inadequate. .. More data need to given

Question 3 |

The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is ?

2.3 m | |

4.6 m | |

7.8 m | |

9.2 m |

Question 3 Explanation:

Let AB be the wall and BC be the ladder.
Then, ACB = 60º and AC = 4.6 m.
AC/BC = cos 60º = 1 2
BC = 2 x AC
= (2 x 4.6) m
= 9.2 m.

Question 4 |

An observer 1.6 m tall is 203 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: ?

21.6 m | |

23.2 m | |

24.72 m | |

None of these |

Question 4 Explanation:

Let AB be the observer and CD be the tower.
Draw BE _|_ CD.
Then, CE = AB = 1.6 m,
BE = AC = 20*squareroot(3) m.
DE/BE = tan 30º = 1/squareroot(3)
DE = 20 m.
CD = CE + DE = (1.6 + 20) m = 21.6 m.

Question 5 |

From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is: ?

149 m | |

156 m | |

173 m | |

200 m |

Question 5 Explanation:

Let AB be the tower.
Then, APB = 30º and AB = 100 m.
AB/AP = tan 30º = 1/squareroot(3)
AP= (AB x squareroot(3)) m= 100*squareroot(3) m = (100 x 1.73)m
=173

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.