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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, angleCAB = 60°, angleDAB = 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, 

\(\tan 45^0 = \dfrac{\text{DB}}{\text{AB}}\\\\1 = \dfrac{\text{DB}}{\text{AB}}\\\\\text{DB} = \text{AB} = 300\sqrt{3}\)

\(\text{Required height = CD = (CB-DB) = }\left(900-300\sqrt{3}\right)\\\\= (900 - 300 \times 1.73) = (900 - 519) = 381\text{ m}\)

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.
 

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.
 

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.
 

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.
           
 

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
 



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°