10th Class Mathematics Blank Heights and Distance

 

Heights and Distances

 

Heights and Distances

 

1. Line of Sight: The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

 

2. Angle of Elevation: The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal when it is above the horizontal level.

 

3. Angle of Depression: The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level.

 

Snap Test

 

 

 

1. The angle of depression of the top and the bottom of a building 50 metres high as observed from the top of a tower are \[\mathbf{30}{}^\circ \] and \[\mathbf{60}{}^\circ \] respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

            (a) 73 m.43.3 m             (b) 90.3 m, 75 m

            (c) 75 m, 41.3 m            (d) 75 m, 43.3 m

            (e) None of these

Ans.     (d)

            Explanation: Let AB be the building and CD be the tower in the figure given below.

 

           

 

          

Let \[BD\text{ }=\text{ }x\text{ }metres\text{ }and~\,CD\,\,=\,\,h\,\,metres\].

            Draw\[AE\parallel BD\].

            Then, \[\angle ~CAE\text{ }=\,\,\angle ~FCA\text{ }=\text{ }30{}^\circ \]

            And   \[~\angle CBD\text{ }=~\angle FCB\text{ }=\text{ }60{}^\circ \]   

            In \[\Delta CAE\]            

\[CE\text{ }=\text{ }h\text{ }-\text{ }50,\,\,\angle \,CAE\text{ }=\text{ }30{}^\circ \text{ }and\text{ }AE\text{ }=\text{ }x\]

 

\[\therefore \,\,\,\,\,\frac{CE}{AE}\,\,=\,\,\tan 30{}^\circ \,\,\,\,\Rightarrow \,\,\,\,\frac{h-50}{x}\,\,\,\,=\,\,\,\frac{1}{\sqrt{3}}\]

 

                        \[\Rightarrow \,\,\,h\,\,-\,\,50\,\,=\,\,\frac{x}{\sqrt{3}}\]      ……. (i)

            In \[\Delta CBD\]

 

            \[CD\text{ }=\text{ }h,\text{ }BD\text{ }=\text{ }x\text{ }and\,\,\angle CBD\text{ }=\text{ }60{}^\circ \]           

 

\[\therefore \,\,\,\frac{CD}{BD}\,\,\,=\,\,\tan \,60{}^\circ \,\,\Rightarrow \,\,=\,\,\frac{h}{x}\,\,=\,\,\sqrt{3}\,\,\Rightarrow \,\,h\,\,\,=\,\,\sqrt{3}\,x\]

 

            Substituting \[h=~\sqrt{3}x\] from (ii) in (i), we get:

 

\[\sqrt{3}x-50\,\,=\,\,\frac{x}{\sqrt{3}}\,\,\Rightarrow \,\,\,3x\,-\,50\sqrt{3}\,\,=\,\,\,x\,\,\,\Rightarrow \,\,2x=50\sqrt{3}\,\,\,\Rightarrow \,\,x=25\sqrt{3}\,\,=\,\,75\].

 

Substituting \[x\text{ }=\text{ }25\sqrt{3}\] in (ii), we get: \[h=\sqrt{3}x\,\,=\sqrt{3}\,\,\times \,\,25\,\sqrt{3}=75.\]

Thus, height of the tower \[CD\text{ }=\text{ }h\text{ }=\text{ }75\] metres and the distance between the building and the tower     \[=\text{ }BD\text{ }=\text{ }x\text{ }=~\,\,25\sqrt{3}\,\,metres\text{ }=\] \[\left( 25\text{ }\times \text{ 1}.732 \right)\text{ }m\text{ }=\text{ }43.3\text{ }m.\]

 

2. A man standing on the deck of a ship, which Is 10 m above the water level, observes the angle of elevation of the top of a hill as \[\mathbf{60}{}^\circ \] and the angel of depression of the base of the hill as\[\mathbf{30}{}^\circ \], Calculate the distance of the hill from the ship and the height of the hill.

            (a) 50 m, 17.32 m                      (b) 16.32 m, 40 m

            (c) 17.32 m, 40 m           (d) 17.32 m, 38 m

            (e) None of these

Ans.     (c)

Explanation: In the following figure, Let AB be the hill and D be the position of the man who is 10 m above the water level EA. Let the height of the hill AB be h metres and the distance CA of the hill from the ship be x metres.

            Draw \[DE\parallel \text{ }CA\]. Then, \[DE\text{ }=\text{ }CA=x\text{ }metres\].

And \[\angle \,BDE\text{ }=\text{ }60{}^\circ ,\,\,\angle \,DAC\text{ }=\,\,\angle \,EDA\text{ }=\text{ }30{}^\circ ,\text{ }AE\text{ }=\text{ }CD\text{ }=\text{ }10\,\,m\text{ }and\text{ }BE\text{ }=\text{ }\left( h\text{ }-\text{ }10 \right)\text{ }m.\]

           

 

          

In\[\Delta BDE\],

                        \[\frac{BE}{DE}=\tan \,60{}^\circ \,\,\Rightarrow \,\,\frac{h-10}{x}\,\,=\,\,\sqrt{3}\]

           

                                                \[\Rightarrow \,\,\,h\,\,\sqrt{3x}\,\,+\,\,10\]            ……. (i)

            In \[\Delta \,DCA\],

 

                        \[\frac{CD}{CA}\,\,=\,\,\tan \,30{}^\circ \,\,\Rightarrow \,\,\,\frac{10}{x}\,\,=\frac{1}{\sqrt{3}}\]

 

                        \[\Rightarrow \,\,\,\,x\,\,=\,\,10\sqrt{3}\]                                        …….. (ii)

 

            Substituting \[\Rightarrow \,\,\,\,x\,\,=\,\,10\sqrt{3}\]from (ii) in (i), we get

\[h\text{ }=\text{ }\left( \text{ }\sqrt{3}\text{ }\times 10\sqrt{3} \right)\,\,+\,\,10\,\,=\,\,30\,\,+\,\,10\,\,=\,\,40.\]

\[\therefore \] Distance of the hill from the ship \[=\,\,\,\,x\,\,\,=\,\,10\sqrt{3}\,\,\text{= }10~\,\,\times \,\,1.732\text{ }=\text{ }17.32\text{ }metres\]

            And the height of the hill \[=\text{ }h\text{ }=\text{ }40\text{ }metres.\]

 

3. Two men on either side of a temple 75 m high observe the angles of elevation of the top of the temple to be \[\mathbf{30}{}^\circ \text{ }\mathbf{and}\text{ }\mathbf{60}{}^\circ \] respectively. Find the distance between the two men.

            (a) 174.2 m                    (b) 173.2 m

            (c) 176.3 m                    (d) 171.2 m

            (e) None of these

Ans.     (b)

Explanation: In the following figure, Let AB be the temple and let C and D be the positions of the two men.

Then, \[AB\text{ }=\text{ }75,\,\,\angle \,ACB\text{ }=\text{ }30{}^\circ ,\,\,\angle \,ADB\text{ }=\text{ }60{}^\circ \].

            Let \[BC\text{ }=\text{ }a\text{ }metres\text{ }and\text{ }BD\text{ }=\text{ }b\] metres.

           

           

         

In \[\Delta \,ABC\]

                                                ^{\[\frac{AB}{BC}\,\,=\,\,\tan \,30{}^\circ \,\,\,\,\Rightarrow \,\,\,\frac{75}{a}\,\,\,=\,\,\frac{1}{\sqrt{3}}\,\,=\,\,\,\,\Rightarrow \,\,a\,\,=\,\,75\sqrt{3}\]}

            In \[\Delta \,ABD\]         

\[\frac{AB}{BC}\,\,\,=\,\,\,\tan \,60{}^\circ \,\,\Rightarrow \,\,\frac{75}{b}\,\,=\,\,\sqrt{3}\,\,\,\,\Rightarrow \,\,\,b\,\,\,=\,\,\frac{75}{\sqrt{3}}\,\,=25\sqrt{3}\]

            \[\therefore \,\] The distance between the two men   \[=\text{ }\left( a\text{ }+\text{ }b \right)m\]

                                                                     = \[(75\sqrt{3}\,\,+\,25\sqrt{3})m\,\]

                                                                     \[=\,\,\,\,100\sqrt{3}\,m\text{ }=\text{ }(100\,\,\times \,~1.732)\text{ }m\text{ }=\text{ }173.2\text{ }m\].

 

4. The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is \[\mathbf{30}{}^\circ \]. On advancing 150 metres towards the foot of the tower, the angle of elevation becomes\[\mathbf{60}{}^\circ \].    Find the tower. \[\mathbf{(Use}\,\,\sqrt{\mathbf{3}}\mathbf{=1}\mathbf{.732)}\]

            (a) 116.8 m                    (b) 126.6 m

            (c) 127.5 m                    (d) 129.9 m

            (e) None of these

Ans.     (d)

Explanation: In the following figure. Let AB be the tower Let P and Q be the points where the angles of elevation of the tower are \[30{}^\circ \text{ }and\text{ }60{}^\circ \] respectively. Then, \[PQ\text{ }=\text{ }150\text{ }m.\]          

            Let the height of the tower \[=\text{ }AB\text{ }=\text{ }h\text{ }metres\]

            Let \[QB\text{ }=\text{ }x\] metres. Then, \[PB\text{ }=\text{ }\left( x\text{ }+\text{ }150 \right)\] metres.

           

 

          

In \[\Delta \,ABP\]                      

            \[\frac{AB}{PB}=\tan \,30{}^\circ \,\,\,\,\Rightarrow \,\,\frac{h}{x+150}\,\,=\,\,\frac{1}{\sqrt{3}}\,\,\,\,\Rightarrow \,\,x+150\,\,\,=\,\,\,\sqrt{3}\,h\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\]                                                     …….. (i)

 

            In \[\Delta \,ABQ\]

            \[\frac{AB}{QB}=\tan \,60{}^\circ \,\,\,\,\Rightarrow \            ,\,\frac{h}{x}\,\,=\,\,\sqrt{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,h=\sqrt{3}x\,\,\,\,\,\Rightarrow \,\,\,\,x=\,\,\frac{h}{\sqrt{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\]                                                                 …….. (ii)

 

            Substituting \[x\,\,=\,\,\frac{h}{\sqrt{3}}\]  from (ii) in (i), we get:   

            \[\frac{h}{\sqrt{3}}+150=\sqrt{3}\,h\,\,\,\,\Rightarrow h+150\sqrt{3}=3h\Rightarrow h=75\sqrt{3}=(75\times 1.732)=129.9\,m\].

 

            Thus, the height of the tower \[=\text{ }h\text{ }=\text{ }129.9\text{ }m.\]

 

5. An aeroplane, when 3000 m high, passes vertically above another plane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are \[\mathbf{60}{}^\circ \text{ }\mathbf{and}\text{ }\mathbf{45}{}^\circ \] respectively. Find the vertical distance between the two aeroplanes.

            (a) 1012 m                     (b) 1268 m

            (c) 927 m                       (d) 1297m

            (e) None of these

Ans.     (b)

            Explanation: In the following figure, let B and C be the positions of the two planes vertically above a point A on         the ground. Let O be the point of observation.

 

           

 

Then, \[AC\text{ }=\text{ }3000\text{ }m,\,\,\angle \,AOB\text{ }=\text{ }45{}^\circ \text{ }and\,\,\angle \,AOC\text{ }=\text{ }60{}^\circ \]

            Let \[OA\text{ }=\text{ }x\text{ }metres\text{ }and\text{ }BC\text{ }=\text{ }h\text{ }metres\].

            Then \[AB\text{ }=\text{ }AC\text{ }-\text{ }BC\text{ }=\text{ }\left( 3000-h \right)\text{ }m\]

            In \[\Delta \,AOB\]

\[\frac{AB}{OA}\,\,=\,\,\tan \,45{}^\circ \,\,\,\,\Rightarrow \,\,\frac{3000-h}{x}=1\,\,\Rightarrow \,x=3000-h\]                                                                                                                  ……. (i)

           

            In \[\Delta \,AOC\]

            \[\frac{AC}{OA}\,\,=\,\,\tan \,60{}^\circ \,\,\,\,\Rightarrow \,\,\,\frac{3000}{x}=\sqrt{3}\,\,\,\,\,\Rightarrow \,\,\,x\,\,=\,\,\frac{3000}{\sqrt{3}}=1000\sqrt{3}\]                                                     …….. (ii)

           

            Equating the values of x in (i) and (ii), we get:

            \[1000\sqrt{3}=3000-h\,\,\,\Rightarrow \,\,h\,\,=\,\,(3000-1000\sqrt{3})m=(3000-1732)m\,\,=\,\,1268m\].

            \[\therefore \]  The vertical distance between the two aeroplanes = 1268 m.