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    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
    A   173 m   
    B   200 m   
    C   273 m   
    D   300 m   
    ANS: C - 273 m

    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.

    2).
    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   43 √units   
    B   8 units   
    C   12 units   
    D   Data inadequate   
    E   None of these   
    ANS: D - Data inadequate

    One of AB, AD and CD must have given. So, the data is inadequate.

    3).
    The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is
    A   2.3 m   
    B   4.6 m   
    C   7.8 m   
    D   9.2 m   
    ANS: D - 9.2 m

    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.

    4).
    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 heights of the tower is
    A   21.6 m   
    B   23.2 m   
    C   24.72 m   
    D   None of these   
    ANS: A - 21.6 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 observer and CD be the tower.

    Draw BE  CD.
    Then, CE = AB = 1.6 m,
          BE = AC = 20√3 m.

    DE = tan 30° = 1
    BE √3

     DE =   20√3   m = 20 m.
    √3

     CD = CE + DE = (1.6 + 20) m = 21.6 m.

    5).
    From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is
    A   149 m   
    B   156 m   
    C   173 m   
    D   200 m   
    ANS: C - 173 m

    Let AB be the tower.

    Then, APB = 30° and AB = 100 m.

    AB = tan 30° = 1
    AP √3

     AP = (AB x √3) m
      = 100√3 m
      = (100 x 1.73) m
      = 173 m.

    6).
    The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree, is
    A   30°   
    B   45°   
    C   60°   
    D   90°   
    ANS: A - 30°

    Let AB be the tree and AC be its shadow.

    Let ACB = .

    Then, AC = √3          cot  = √3
    AB

      = 30°.