NCERT Solutions for Class 10 Maths Chapter 9 Some Applications Of Trigonometry Ex 9.1

1. A circus artist is climbing a \[20{\text{ m}}\] long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is \[30^\circ \].

Ans:

Given: A circus performer is climbing a $20\;{\text{m}}$ rope that runs from the top of a vertical pole to the ground, firmly stretched and knotted. 

To find: If the rope makes a \[30^\circ \] angle with the ground level, calculate the height of the pole.

It can be observed from the figure that AB is the pole. 

In $\vartriangle {\text{ABC}}$,

$\dfrac{{{\text{AB}}}}{{{\text{AC}}}} = \sin {30^\circ }$

$\dfrac{{{\text{AB}}}}{{20}} = \dfrac{1}{2}$

${\text{AB}} = \dfrac{{20}}{2} = 10$

Therefore, the height of the pole is $10\;\;{\text{m}}$.

2. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle \[30^\circ \] with it. The distance between the feet of the tree to the point where the top touches the ground is $8{\text{ m}}$. Find the height of the tree.

Ans:

Given: A storm causes a tree to shatter, and the fractured section bends such that the top of the tree meets the ground at a \[30^\circ \] angle. The distance from the tree's foot to where the top touches the earth is $8$ meters. 

To find: Determine the tree's height.

The original tree was Let AC. It was split in two due to the storm. The broken portion A'B is at a \[30^\circ \]angle to the ground.

In $\Delta {{\text{A}}^\prime }{\text{BC}}$

$\dfrac{{{\text{BC}}}}{{{{\text{A}}^\prime }{\text{C}}}} = \tan {30^\circ }$

$\dfrac{{{\text{BC}}}}{8} = \dfrac{1}{{\sqrt 3 }}$

${\text{BC}} = \left( {\dfrac{8}{{\sqrt 3 }}} \right){\text{m}}$

$\dfrac{{{{\text{A}}^\prime }{\text{C}}}}{{{{\text{A}}^\prime }{\text{B}}}} = \cos {30^\circ }$

$\dfrac{8}{{\;{{\text{A}}^\prime }{\text{B}}}} = \dfrac{{\sqrt 3 }}{2}$

$\;{{\text{A}}^\prime }{\text{B}} = \left( {\dfrac{{16}}{{\sqrt 3 }}} \right){\text{m}}$

Height of tree $ = {{\text{A}}^\prime }{\text{B}} + {\text{BC}}$

$ = \left( {\dfrac{{16}}{{\sqrt 3 }} + \dfrac{8}{{\sqrt 3 }}} \right){\text{m}} = \dfrac{{24}}{{\sqrt 3 }}\;{\text{m}}$

$ = 8\sqrt 3 \;{\text{m}}$

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of $5$ years, she prefers to have a slide whose top is at a height of $1.5\,{\text{m}}$ , and is inclined at an angle of \[30^\circ \] to the ground, whereas for the elder children she wants to have a steep slide at a height of $3\;{\text{m}}$, and inclined at an angle of \[60^\circ \] to the ground. What should be the length of the slide in each case?

Ans:

Given: In a park, a contractor intends to construct two slides for the kids to enjoy. She likes a slide with a top height of $1.5\,{\text{m}}$ and an angle of \[30^\circ \] to the ground for children under the age of $5$, and a steep slide with a top height of $3\;{\text{m}}$ and an angle of \[60^\circ \] to the ground for older children. 

To find: In each scenario, what should the length of the slide be?

It can be noted that AC and PR are the slides for younger and elder children.

In $\vartriangle {\text{ABC}}$,

$\dfrac{{{\text{AB}}}}{{{\text{AC}}}} = \sin {30^\circ }$

$\dfrac{{1.5}}{{{\text{AC}}}} = \dfrac{1}{2}$

${\text{AC}} = 3\;{\text{m}}$

In $\Delta {\text{PQR}}$,

$\dfrac{{{\text{PQ}}}}{{{\text{PR}}}} = \sin 60$

$\dfrac{3}{{PR}} = \dfrac{{\sqrt 3 }}{2}$

${\text{PR}} = \dfrac{6}{{\sqrt 3 }} = 2\sqrt 3 \;{\text{m}}$

Therefore, the lengths of these slides are $3\;{\text{m}}$ and $2\sqrt 3 \;{\text{m}}$.

4. The angle of elevation of the top of a tower from a point on the ground, which is $30{\text{ m}}$ away from the foot of the tower is \[30^\circ \]. Find the height of the tower.

Ans:

Given: The angle of elevation of a tower's top from a point on the ground $30$ metres distant from the tower's foot is \[30^\circ \]. 

To find: Determine the tower's height.

Let AB be the tower and the angle of elevation from point C (on ground) is \[30^\circ \]. In \[\Delta ABC\],

In \[\Delta ABC\], $\dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \tan {30^\circ }$

$\dfrac{{{\text{AB}}}}{{30}} = \dfrac{1}{{\sqrt 3 }}$

${\text{AB}} = \dfrac{{30}}{{\sqrt 3 }} = 10\sqrt 3 \;{\text{m}}$

Therefore, the height of the tower is $10\sqrt 3 \;{\text{m}}$.

5. A kite is flying at a height of $60\;{\text{m}}$ above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is \[60^\circ \]. Find the length of the string, assuming that there is no slack in the string.

Ans:

Given: At a height of $60\;{\text{m}}$ above the earth, a kite is flying. The kite's line is temporarily tethered to a location on the ground. The string is at a \[60^\circ \]angle to the ground. 

To find: Determine the string's length, assuming there is no slack in the string.

Let K be the kite and the string is tied to point P on the ground. In \[\Delta KLP\],

In $\Delta {\text{KLP}}$,

$\dfrac{{{\text{KL}}}}{{{\text{KP}}}} = \sin {60^\circ }$

$\dfrac{{60}}{{{\text{KP}}}} = \dfrac{{\sqrt 3 }}{2}$

${\text{KP}} = \dfrac{{120}}{{\sqrt 3 }} = 40\sqrt 3 \;{\text{m}}$

Hence, the length of the string is $40\sqrt 3 \;{\text{m}}$.

6. A 1.5 m tall boy is standing at some distance from a $30\,{\text{m}}$ tall building. The angle of elevation from his eyes to the top of the building increases from \[30^\circ \] to \[60^\circ \] as he walks towards the building. Find the distance he walked towards the building.

Ans:

Given: A 1.5-meter-tall child stands at a safe distance from a $30\,{\text{m}}$ tall structure. As he goes towards the structure, the angle of elevation from his eyes to the top of the building grows from \[30^\circ \] to \[60^\circ \]. 

To find: Calculate how far he went towards the building.

Let the boy was standing at point S initially. He walked towards the building and reached point T.

${\text{PR}} = {\text{PQ}} - {\text{RQ}}$

$ = (30 - 1.5){\text{m}} = 28.5\;{\text{m}} = \dfrac{{57}}{2}\;{\text{m}}$

In $\vartriangle {\text{PAR}}$,

$\dfrac{{{\text{PR}}}}{{{\text{AR}}}} = \tan {30^\circ }$

$\dfrac{{57}}{{2{\text{AR}}}} = \dfrac{1}{{\sqrt 3 }}$

${\text{AR}} = \left( {\dfrac{{57}}{2}\sqrt 3 } \right){\text{m}}$

In $\vartriangle {\text{PRB}}$

$\dfrac{{{\text{PR}}}}{{{\text{BR}}}} = \tan 60$

$\dfrac{{57}}{{2{\text{BR}}}} = \sqrt 3 $

${\text{BR}} = \dfrac{{57}}{{2\sqrt 3 }} = \left( {\dfrac{{19\sqrt 3 }}{2}} \right){\text{m}}$

${\text{ST}} = {\text{AB}}$

$ = {\text{AR}} - {\text{BR}} = \left( {\dfrac{{57\sqrt 3 }}{2} - \dfrac{{19\sqrt 3 }}{2}} \right){\text{m}}$

$ = \left( {\dfrac{{38\sqrt 3 }}{2}} \right){\text{m}} = 19\sqrt 3 \;{\text{m}}$

Hence, he walked $19\sqrt 3 \;{\text{m}}$ towards the building.

7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20\,{\text{m}}$ high building are \[45^\circ \] and \[60^\circ \] respectively. Find the height of the tower.

Ans:

Given: The angles of elevation of the bottom and top of a transmission tower installed at the top of a $20\,{\text{m}}$ high structure are \[45^\circ \] and \[60^\circ \], respectively, from a point on the ground. 

To find: Determine the tower's height.

Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.

In $\vartriangle {\text{BCD}}$,

$\dfrac{{{\text{BC}}}}{{{\text{CD}}}} = \tan {45^\circ }$

$\dfrac{{20}}{{{\text{CD}}}} = 1$

${\text{CD}} = 20\;{\text{m}}$

In $\vartriangle {\text{ACD}}$,

$\dfrac{{{\text{AC}}}}{{{\text{CD}}}} = \tan {60^\circ }$

$\dfrac{{{\text{AB}} + {\text{BC}}}}{{{\text{CD}}}} = \sqrt 3 $

$\dfrac{{{\text{AB}} + 20}}{{{\text{CD}}}} = \sqrt 3 $

${\text{AB}} = (20\sqrt 3  - 20){\text{m}}$

$ = 20(\sqrt 3  - 1){\text{m}}$

Therefore, the height of the transmission tower is $20(\sqrt 3  - 1){\text{m}}$

8. A statue, $1.6\,{\text{m}}$ tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is \[60^\circ \] and from the same point the angle of elevation of the top of the pedestal is \[45^\circ \]. Find the height of the pedestal.

Ans:

Given: A $1.6\,{\text{m}}$ tall statue sits on top of a pedestal; the angle of elevation of the top of the statue is \[60^\circ \] from a point on the ground, and the angle of elevation of the top of the pedestal is \[45^\circ \] from the same point. 

To find: Determine the pedestal's height.

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.

In $\vartriangle {\text{BCD}}$,

$\dfrac{{{\text{BC}}}}{{{\text{CD}}}} = \tan {45^\circ }$

$\dfrac{{BC}}{{CD}} = 1$

${\text{BC}} = {\text{CD}}$

In $\vartriangle {\text{ACD}}$,

$\dfrac{{{\text{AB}} + {\text{BC}}}}{{{\text{CD}}}} = \tan 60$

$\dfrac{{{\text{AB}} + {\text{BC}}}}{{{\text{CD}}}} = \sqrt 3 $

$1.6 + {\text{BC}} = {\text{BC}}\sqrt 3 \quad [{\text{AsCD}} = {\text{BC}}]$

${\text{BC}}(\sqrt 3  - 1) = 1.6$

${\text{BC}} = \dfrac{{(1.6)(\sqrt 3  + 1)}}{{(\sqrt 3  - 1)(\sqrt 3  + 1)}}$

(By Rationalization)

$ = \dfrac{{1.6(\sqrt 3  + 1)}}{{{{(\sqrt 3 )}^2} - {{(1)}^2}}}$

$ = \dfrac{{1.6(\sqrt 3  + 1)}}{2} = 0.8(\sqrt 3  + 1)$

Therefore, the height of the pedestal is $0.8(\sqrt 3  + 1)\;{\text{m}}$

9. The angle of elevation of the top of a building from the foot of the tower is \[30^\circ \] and the angle of elevation of the top of the tower from the foot of the building is \[60^\circ \]. If the tower is $50{\text{ m}}$ high, find the height of the building.

Answer :

Given: The angle of elevation of a building's top from the foot of the tower is \[30^\circ \], while the angle of elevation of a tower from the foot of the building is \[60^\circ \]. 

To find: Find the height of the building if the tower is $50{\text{ m}}$ tall.

Let AB be the building and CD be the tower. 

In $\Delta {\text{CDB}}$,

$\dfrac{{{\text{CD}}}}{{{\text{BD}}}} = \tan {60^\circ }$

$\dfrac{{50}}{{{\text{BD}}}} = \sqrt 3 $

${\text{BD}} = \dfrac{{50}}{{\sqrt 3 }}$

In $\vartriangle {\text{ABD}}$,

$\dfrac{{{\text{AB}}}}{{{\text{BD}}}} = \tan {30^\circ }$

${\text{AB}} = \dfrac{{50}}{{\sqrt 3 }} \times \dfrac{1}{{\sqrt 3 }} = \dfrac{{50}}{3} = 16\dfrac{2}{3}$

Therefore, the height of the building is $16\dfrac{2}{3}\;{\text{m}}$.

10. Two poles of equal heights are standing opposite each other on either side of the road, which is $80{\text{ m}}$ wide. From a point between them on the road, the angles of elevation of the top of the poles are \[60^\circ \] and \[30^\circ \], respectively. Find the height of poles and the distance of the point from the poles.

Ans:

Given: On either side of the $80{\text{ m}}$ wide road, two poles of equal height stand opposite each other. The angles of elevation of the tops of the poles are \[60^\circ \] and \[30^\circ \], respectively, from a point on the route between them. 

To find: Determine the height of the poles and the distance between them.

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

In $\vartriangle {\text{ABO}}$,

$\dfrac{{{\text{AB}}}}{{{\text{BO}}}} = \tan 60$

$\dfrac{{{\text{AB}}}}{{{\text{BO}}}} = \sqrt 3 $

${\text{BO}} = \dfrac{{{\text{AB}}}}{{\sqrt 3 }}$

In $\vartriangle {\text{CDO}}$,

$\dfrac{{CD}}{{DO}} = \tan {30^\circ }$

$\dfrac{{{\text{CD}}}}{{80 - {\text{BO}}}} = \dfrac{1}{{\sqrt 3 }}$

${\text{CD}}\sqrt 3  = 80 - {\text{BO}}$

$CD\sqrt 3  = 80 - \dfrac{{AB}}{{\sqrt 3 }}$

${\text{CD}}\sqrt 3  + \dfrac{{{\text{AB}}}}{{\sqrt 3 }} = 80$

Since the poles are of equal heights, ${\text{CD}} = {\text{AB}}$

${\text{CD}}\left[ {\sqrt 3  + \dfrac{1}{{\sqrt 3 }}} \right] = 80$

${\text{CD}}\left( {\dfrac{{3 + 1}}{{\sqrt 3 }}} \right) = 80$

${\text{CD}} = 20\sqrt 3 \;{\text{m}}$

${\text{BO}} = \dfrac{{{\text{AB}}}}{{\sqrt 3 }} = \dfrac{{{\text{CD}}}}{{\sqrt 3 }} = \left( {\dfrac{{20\sqrt 3 }}{{\sqrt 3 }}} \right){\text{m}} = 20\;{\text{m}}$

${\text{DO}} = {\text{BD}} - {\text{BO}} = (80 - 20){\text{m}} = 60\;{\text{m}}$

Therefore, the height of poles is $20\sqrt 3 $ and the point is $20\;\;{\text{m}}$ and $60\;\;{\text{m}}$ far from the D

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is \[60^\circ \]. From another point $20\;\;{\text{m}}$ away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is \[30^\circ \]. Find the height of the tower and the width of the canal.

Ans:

Given: On the edge of a canal, a TV tower stands vertically. The angle of elevation of the top of the tower from a location on the other bank exactly opposite the tower is \[60^\circ \]. The angle of elevation of the top of the tower is \[30^\circ \] measured from another point $20\;\;{\text{m}}$ distant on the line connecting this point to the foot of the tower. 

To find: Calculate the tower's height and the canal's width.

In $\Delta ABC$

$\dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \tan {60^\circ }$

$\dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \sqrt 3 $

${\text{BC}} = \dfrac{{{\text{AB}}}}{{\sqrt 3 }}$

In $\vartriangle {\text{ABD}}$,

$\dfrac{{{\text{AB}}}}{{{\text{BD}}}} = \tan 30$

$\dfrac{{{\text{AB}}}}{{{\text{BC}} + {\text{CD}}}} = \dfrac{1}{{\sqrt 3 }}$

$\dfrac{{{\text{AB}}}}{{\dfrac{{{\text{AB}}}}{{\sqrt 3 }} + 20}} = \dfrac{1}{{\sqrt 3 }}$

$\dfrac{{{\text{AB}}\sqrt 3 }}{{{\text{AB}} + 20\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}$

$3{\text{AB}} = {\text{AB}} + 20\sqrt 3 $

$2{\text{AB}} = 20\sqrt 3 $

${\text{AB}} = 10\sqrt 3 \;{\text{m}}$

${\text{BC}} = \dfrac{{{\text{AB}}}}{{\sqrt 3 }} = \left( {\dfrac{{10\sqrt 3 }}{{\sqrt 3 }}} \right){\text{m}} = 10\;{\text{m}}$

Therefore, the height of the tower is $10\sqrt 3 \;\;{\text{m}}$ and the width of the canal is $10\;\;{\text{m}}$.

12. From the top of a $7\;{\text{m}}$ high building, the angle of elevation of the top of a cable tower is \[60^\circ \] and the angle of depression of its foot is \[45^\circ \]. Determine the height of the tower.

Ans:

Given: From the top of a $7\;{\text{m}}$ high building, the angle of elevation of the top of a cable tower is \[60^\circ \] and the angle of depression of its foot is \[45^\circ \]. 

To find: Determine the height of the tower.

Let ${\text{AB}}$ be a building and ${\text{CD}}$ be a cable tower. In $\vartriangle {\text{ABD}}$,

$\dfrac{{{\text{AB}}}}{{{\text{BD}}}} = \tan 45$

$\dfrac{7}{{{\text{BD}}}} = 1 \Rightarrow {\text{BD}} = 7\;{\text{m}}$

In $\vartriangle {\text{ACE}}$,

${\text{AE}} = {\text{BD}} = 7\;{\text{m}}$

$\dfrac{{{\text{CE}}}}{{{\text{AE}}}} = \tan {60^\circ }$

$\dfrac{{{\text{CE}}}}{7} = \sqrt 3 $

${\text{CE}} = 7\sqrt 3 \;{\text{m}}$

${\text{CD}} = {\text{CE}} + {\text{ED}} = (7\sqrt 3  + 7){\text{m}}$

$ = 7(\sqrt 3  + 1){\text{m}}$

Therefore, the height of the cable tower is $(7\sqrt 3  + 7)\,{\text{m}}$.

13. As observed from the top of a $75\;{\text{m}}$ high lighthouse from the sea-level, the angles of depression of two ships are \[30^\circ \] and \[45^\circ \]. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Ans:

Given: The angles of depression of two ships are \[30^\circ \] and \[45^\circ \], respectively, as seen from the top of a $75\;{\text{m}}$ high lighthouse from sea level.

To find: Find the distance between the two ships if one is precisely behind the other on the same side of the lighthouse.

$\text { Let the distance between the ships be } \mathrm{x} \mathrm{m} \text {. }$

Then in right $\triangle \mathrm{ADB}, \tan 30^{\circ}=\dfrac{\mathrm{AB}}{\mathrm{BD}}$

$\Rightarrow \quad 1=\dfrac{75}{y} \quad \Rightarrow \quad y=75 \mathrm{~m}$

In right $\triangle \mathrm{ADB}, \tan 30^{\circ}=\dfrac{\mathrm{AB}}{\mathrm{BD}}$

$\Rightarrow \quad \dfrac{1}{\sqrt{3}}=\dfrac{75}{x+y}$ $\Rightarrow x+75=75 \sqrt{3} \quad \quad [$ From equation (i) $] $

$\Rightarrow \quad x=75 \sqrt{3}-75=75(\sqrt{3}-1)$

Hence, the distance between the two ships is $\mathbf{7 5}(\sqrt{\mathbf{3}}-\mathbf{1}) \mathbf{m}$

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2\;{\text{m}}$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is \[60^\circ \]. After some time, the angle of elevation reduces to \[30^\circ \]. Find the distance travelled by the balloon during the interval.

Ans:

Given: A 1.2 m tall girl notices a balloon floating horizontally with the wind at a height of $88.2\;{\text{m}}$ above the earth. At any given time, the angle of elevation of the balloon from the girl's eyes is \[60^\circ \]. After a while, the angle of elevation drops to \[30^\circ \]

To find: Calculate the balloon's distance travelled throughout the timeframe.

Given that $1.2 \mathrm{~m}$ tall girl sees a balloon

So, $A G=1.2 \mathrm{~m}$

Also, AG \& BF are parallel $B F=A G=1.2 \mathrm{~m}$

And the balloon is at a height of $88.2 \mathrm{~m}$

So, EF $=88.2$

Girl sees balloon first at $60^{\circ}$

So, $\angle E A B=60^{\circ}$

After travelling, the angle of elevation becomes $30^{\circ}$

So, $\angle D A C=30^{\circ}$

Distance travelled by the balloon $=\mathrm{BC}$

We have to find BC

Now,

B E = E F - B F 

B E = 88.2 - 1.2 m

B E = 87 m

$\mathrm{Also}, \mathrm{BE}=\mathrm{DC}=87 \mathrm{~m}$

Here,

$\angle \mathrm{ABE}=90^{\circ} \& \angle \mathrm{ACD}=90^{\circ}$

In right angle triangle EBA

$\tan \mathrm{A}=\dfrac{\text { Side opposite to angle } \mathrm{A}}{\text { Side adjacent to angle } \mathrm{A}} $

$ tan \mathrm{A}=\dfrac{B E}{A B}$

$\tan 60^{\circ}=\dfrac{B E}{A B}$

$\sqrt{3}=\dfrac{87}{A B}$

$A B=\dfrac{87}{\sqrt{3}} \mathrm{~m}$

In right angle triangle DAC

$\tan \mathrm{A}=\dfrac{\text { Side opposite to angle } \mathrm{A}}{\text { Side adjacent to angle } \mathrm{A}} $

$\tan \mathrm{A}=\dfrac{C D}{A C}$

$\tan 30^{\circ}=\dfrac{C D}{A C}$

$\dfrac{1}{\sqrt{3}}=\dfrac{87}{A C}$

$\mathrm{AC}=87 \sqrt{3} \mathrm{~m}$

$A C=A B+B C$

$87 \sqrt{3}=\dfrac{87}{\sqrt{3}}+B C$

$87 \sqrt{3}-\dfrac{87}{\sqrt{3}}=B C$

$B C=87 \sqrt{3}-\dfrac{87}{\sqrt{3}}$

$B C=87\left(\sqrt{3}-\dfrac{1}{\sqrt{3}}\right)$

$B C=87\left(\dfrac{\sqrt{3} \sqrt{3}-1}{\sqrt{3}}\right)$

$B C=87\left(\dfrac{3-1}{\sqrt{3}}\right)$

$B C=87\left(\dfrac{2}{\sqrt{3}}\right)$

$B C=\dfrac{87 \times 2}{\sqrt{3}}$

Multiply $\sqrt{3}$ in numerator and denominator

$\mathrm{BC}=\dfrac{87 \times 2}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} $

$\mathrm{BC}=\dfrac{87 \times 2 \times \sqrt{3}}{3} $

$\mathrm{BC}=29 \times 2 \times \sqrt{3} $

$\mathrm{BC}=58 \sqrt{3}$

Hence, distance travelled by balloon $=58 \sqrt{3} \mathrm{~m}$

15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of \[30^\circ \], which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be \[60^\circ \]. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

Let the height of the tower DC be $h \mathrm{~m}$ and the speed of the car be $\mathrm{x} \mathrm{m} / \mathrm{s}$. Then the distance covered by the car in $6 \mathrm{~s}$ will be $6 \mathrm{x} \mathrm{m}$.

Also let the time taken by car to move from $B$ to $C$ be $t \mathrm{~s}$.

Then the distance BC will be $x t \mathrm{~m}$

In right $\triangle \mathrm{DCB}$,

$\tan 60^{\circ}=\dfrac{\mathrm{DC}}{\mathrm{BC}} \quad \Rightarrow \quad \sqrt{3}=\dfrac{h}{t x} $

$\Rightarrow \quad h=\sqrt{3} t x$

In right $\triangle \mathrm{ACD}$,

tan 30° = $ \frac{DC}{AC} $

$\Rightarrow \quad \dfrac{1}{\sqrt{3}}=\dfrac{h}{6 x+t x} $ 

$\Rightarrow  6 x+t x =\sqrt{3} h $

 $\Rightarrow x(6+t) =\sqrt{3} \times \sqrt{3} t x $

$\Rightarrow  6+t =3 t \quad \Rightarrow \quad 2 t=6 $

$\Rightarrow  \quad \text { [From (i)] }$

$ \Rightarrow  t =3 \mathrm{~s} $

Hence, the required time taken by car is $\mathbf{3} \mathbf{~ s .}$

Conclusion

In conclusion, Chapter 9 of Class 10 Maths, Some Applications of Trigonometry, provides students with essential skills for solving practical problems involving heights and distances. Ex 9.1 class 10 lays the foundation by introducing the concepts of angles of elevation and depression and the application of trigonometric ratios in real-life scenarios. Over the years, questions from Class 10 Maths Ex 9.1 have frequently appeared in board exams, emphasizing its importance on calculating heights and distances using the angles of elevation and depression. These questions test students' understanding of basic trigonometric applications and their ability to apply these concepts to solve real-world problems.

Other Study Materials of CBSE Class 10 Maths Chapter 9

Chapter-Specific NCERT Solutions for Class 10 Maths

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