## Aptitude question and answer on Height and distance

 1 An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60° and 45° respectively. Approximately, how many meters higher is the one than the other? A.    381 m B.    169 m C.    254 m D.    211 m Answer : A.  381 m Explanation: Let C and D be the position of the aeroplanes. Given that CB = 900 m, CAB = 60°, DAB = 45°  From the right $\bigtriangleup$ ABC,  $\tan 60? = \dfrac{\text{CB}}{\text{AB}}\\\\\sqrt{3} = \dfrac{900}{\text{AB}}\\\\ \text{AB} = \dfrac{900}{\sqrt{3}} = \dfrac{900 \times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} = \dfrac{900\sqrt{3}}{3} = 300\sqrt{3}$ From the right $\bigtriangleup$ ABD,  $\tan 45^0 = \dfrac{\text{DB}}{\text{AB}}\\\\1 = \dfrac{\text{DB}}{\text{AB}}\\\\\text{DB} = \text{AB} = 300\sqrt{3}$ $\text{Required height = CD = (CB-DB) = }\left(900-300\sqrt{3}\right)\\\\= (900 - 300 \times 1.73) = (900 - 519) = 381\text{ m}$ Discuss 2 The angle of elevation of the top of a tower at a point on the ground is 30°. On walking 24 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower. A.    20.56 m B.    20.66 m C.    20.76 m D.    20.86 m Answer : C.  20.76 m Explanation: Let AB be the tower and C and D be the points of observation. Then, AB/AD = tan 60° = $\sqrt{3}$          => AD = AB/$\sqrt{3}$  = h/$\sqrt{3}$             AB/AC = tan 30° = 1/$\sqrt{3}$ => AC=AB x $\sqrt{3}$ = h$\sqrt{3}$   CD = (AC-AD) = (h$\sqrt{3}$-$h \over\sqrt{3}$)   h$\sqrt{3}$-$h \over\sqrt{3}$ = 24  => h=12$\sqrt{3}$ = (12$\times$1.73) = 20.76 Hence, the height of the tower is 20.76 m.   Hence, the height of the tower is 20.76 m.   Discuss 3 A man standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retires 36 m from the bank, he finds the angle to be 30°. Find the breadth of the river. A.    15 m B.    16 m C.    17 m D.    18 m Answer : D.  18 m Explanation: Let AB be the tree and AC be the river. Let C and D be the two positions of the man. Then,             ACB=60°, ADB=30° and CD=36 m.             Let AB=h metres and AC=x metres.             Then, AD=(36+x)metres.             AB/AD=tan 30°=1/$\sqrt{3}$                        =>        h/(36+x)=1/$\sqrt{3}$             h=(36+x)/ $\sqrt{3}$    .............................(1)             AB/AC=tan 60°=$\sqrt{3}$              =>        h/x=$\sqrt{3}$             h=$\sqrt{3}$x             .............................(2)             From (i) and (ii), we get:             (36+x)/ $\sqrt{3}$ = $\sqrt{3}$ x       =>        x=18 m.               So, the breadth of the river = 18 m.   Discuss 4 A man on the top of a tower, standing on the seashore finds that a boat coming towards him takes 10 minutes for the angle of depression to change from 30° to 60°. Find the time taken by the boat to reach the shore from this position. A.    5 minutes B.    6 minutes C.    7 minutes D.    8 minutes Answer : A.  5 minutes Explanation: Let AB be the tower and C and D be the two positions of the boat.               Let AB=h, CD=x and AD=y.               h/y=tan 60°=$\sqrt{3}$                      =>        y=h/$\sqrt{3}$               h/(x+y)=tan 30° = 1/$\sqrt{3}$   =>        x+y=$\sqrt{3}$h               x=(x+y)-y = ($\sqrt{3}$h-h/$\sqrt{3}$)=2h/$\sqrt{3}$               Now, 2h/$\sqrt{3}$ is covered in 10 min.                         h/$\sqrt{3}$ will be covered in (10$\times$($\sqrt{3}$/2h)$\times$(h/$\sqrt{3}$))=5 min               Hence, required time = 5 minutes.   Discuss 5 There are two temples, one on each bank of a river, just opposite to each other. One temple is 54 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width of the river and the height of the other temple. A.    16 m B.    17 m C.    18 m D.    19 m Answer : C.  18 m Explanation: Let AB and CD be the two temples and AC be the river.             Then, AB =  54 m.             Let AC = x metres and CD=h metres.               ACB=60°, EDB=30°             AB/AC=tan 60°=$\sqrt{3}$             AC=AB/$\sqrt{3}$=54/$\sqrt{3}$=(54/$\sqrt{3}$ $\times$$\sqrt{3}$/$\sqrt{3}$)=18m             DE=AC=18$\sqrt{3}$             BE/DE=tan 30°=1/$\sqrt{3}$             BE=(18$\sqrt{3}$$\times$1/$\sqrt{3}$)=18 m             CD=AE=AB-BE=(54-18) m = 36 m.             So, Width of the river = AC = 18$\sqrt{3}$ m=18 $\times$1.73 m=31.14m             Height of the other temple = CD= 18 m.               Discuss 6 A ladder leaning against a wall makes an angle of 60° with the ground. If the length of the ladder is 19 m, find the distance of the foot of the ladder from the wall. A.    9.3 m B.    9.4 m C.    9.5 m D.    9.6 m Answer : C.  9.5 m Explanation: Let AB be the wall and BC be the ladder.             Then, ACB = 60° and BC = 19 m.             Let AC = x metres             AC/BC = cos 60° =>    $x\over19$ = ½   =>  x=19/2 = 9.5             $\dot{..}$ Distance of the foot of the ladder from the wall = 9.5 m   Discuss 7 If the height of a pole is $2 \sqrt{3}$ metres and the length of its shadow is 2 metres, find the angle of elevation of the sun. A.    40° B.    50° C.    60° D.    70° Answer : C.  60° Explanation: Let AB be the pole and AC be its shadow.             Let angle of elevation, ACB=q.             Then, AB = $2 \sqrt{3}$ m , AC = 2 m.                         $\tan \theta$ = AB/AC = $2 \sqrt{3}$/2 = $\sqrt{3}$ => $\theta$ = 60°                         So, the angle of elevation is 60°   Discuss