NCERT Solutions for Class 10 Maths Chapter 9 - Exercise 9.1

Exercise No: 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 $\mathbf{3} \mathbf{~ s .}$

16. The angles of elevation of the top of a tower from two points at a distance of $4\;{\text{m}}$ and $9\;{\text{m}}$ from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $6\;{\text{m}}$.

Ans:

Let the height of the tower, $\mathrm{PQ}=\mathrm{h} \mathrm{m}$.

A Q=4 m, B Q=9 m 

$\angle P A Q=\theta \angle P B Q=90-\theta $

$\therefore \quad \text { In } \perp \Delta P A Q, \quad \dfrac{P Q}{A Q}=\tan \theta $

$\dfrac{h}{4}=\tan \theta \quad \ldots \ldots \text { (i) }$

$\text { In } \perp \Delta P B Q, \quad \dfrac{P Q}{B Q}=\tan (90-\theta) $

$\dfrac{h}{9}=\cot \theta $

$\dfrac{h}{9}=\dfrac{1}{\tan \theta} \quad \cdots \ldots \text { (ii) }$

From eqn. (i) and eqn. (ii),

$\dfrac{\mathrm{h}}{4} \times \dfrac{\mathrm{h}}{9}=\tan \theta \times \dfrac{1}{\tan \theta} $

$\dfrac{\mathrm{h}}{4} \times \dfrac{\mathrm{h}}{9}=1 $

$\therefore \mathrm{h}^{2}=36 $

$\therefore \mathrm{h}=6 \mathrm{~m}$

$\therefore$ Height of tower, $\mathrm{h}=6 \mathrm{~m}$.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1

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NCERT Solutions for Class 10 Maths PDFs (Chapter-wise)

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability