NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry – Here are all the NCERT solutions for Class 10 Maths Chapter 9. This solution contains questions, answers, images, explanations of the complete Chapter 9 titled Some Applications of Trigonometry of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 9 Some Applications of Trigonometry. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry in one place.

## NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry

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 Class 10 Subject Maths Book Mathematics Chapter Number 9 Chapter Name Some Applications of Trigonometry

### NCERT Solutions Class 10 Maths chapter 9 Some Applications of Trigonometry

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### Some Applications of Trigonometry

Q.1: A circus artist is climbing a  20 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 : $$\begin{array}{l} \text{It can be observed from the figure that AB is the pole.}{\text { In } \triangle A B C,} \\ {\frac{A B}{A C}=\sin 30^{\circ}} \\ {\frac{A B}{20}=\frac{1}{2}} \\ {A B=\frac{20}{2}=10} \\ {\text { Therefore, the height of the pole is } 10 \mathrm{m} \text { . }}\end{array}$$
Q.2: A tree breaks due to 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 foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Ans :
Let AC was the original tree. Due to storm, it was broken into parts. The broken part
$$\mathrm{A}^{\prime} \mathrm{B} \text { is making } 30^{\circ}$$ with the ground.
$$\begin{array}{l}{\text { In } \Delta \mathrm{A}^{\prime} \mathrm{BC}} \\ {\frac{\mathrm{BC}}{\mathrm{A}^{\prime} \mathrm{C}}=\tan 30^{\circ}} \\ {\frac{\mathrm{BC}}{8}=\frac{1}{\sqrt{3}}} \\ {\mathrm{BC}=\left(\frac{8}{\sqrt{3}}\right) \mathrm{m}}\end{array}$$
$$\begin{array}{l}{\frac{\mathrm{A}^{\prime} \mathrm{C}}{\mathrm{A}^{\prime} \mathrm{B}}=\cos 30^{\circ}} \\ {\frac{8}{\mathrm{A}^{\prime} \mathrm{B}}=\frac{\sqrt{3}}{2}} \\ {\mathrm{A}^{\prime} \mathrm{B}=\left(\frac{16}{\sqrt{3}}\right) \mathrm{m}} \\ {\text { Height of tree }=\mathrm{A}^{\prime} \mathrm{B}+\mathrm{BC}}\end{array}$$
\begin{aligned} &=\left(\frac{16}{\sqrt{3}}+\frac{8}{\sqrt{3}}\right) \mathrm{m}=\frac{24}{\sqrt{3}} \mathrm{m} \\ &=8 \sqrt{3} \mathrm{m} \end{aligned}
Hence, the height of the tree is $$8 \sqrt{3} \mathrm{m}$$. 
Q.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 m, and is inclined at an angle of 30$$^{\circ}$$ to the ground, where as for the elder children she wants to have a steep side at a height of 3 m, and inclined at an angle of 60$$^{\circ}$$ to the ground. What should be the length of the slide in each case?
Ans : It can be observed that AC and PR are the slides for younger and elder children respectively.

$$\begin{array}{l}{\text { In } \triangle A B C,} \\ {\frac{A B}{A C}=\sin 30^{\circ}} \\ {\frac{1.5}{A C}=\frac{1}{2}} \\ {A C=3 \mathrm{m}}\end{array}$$

$$\begin{array}{l}{\text { In } \Delta \mathrm{PQR}_{,}} \\ {\frac{\mathrm{PQ}}{\mathrm{PR}}=\sin 60} \\ {\frac{3}{\mathrm{PR}}=\frac{\sqrt{3}}{2}} \\ {\mathrm{PR}=\frac{6}{\sqrt{3}}=2 \sqrt{3} \mathrm{m}}\end{array}$$
Therefore, the length of these slides are 3 m and $$2 \sqrt{3} \mathrm{m}$$. 
Q.4: The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30$$^{\circ}$$. Find the height of the tower.
Ans :
Let AB be the tower and the angle of elevation from point C(on ground) is $$30^{\circ}$$.
In $$\triangle A B C$$,
$$\begin{array}{l}{\frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ}} \\ {\frac{\mathrm{AB}}{30}=\frac{1}{\sqrt{3}}} \\ {\mathrm{AB}=\frac{30}{\sqrt{3}}=10 \sqrt{3} \mathrm{m}}\end{array}$$
Therefore, the height of the tower is $$10 \sqrt{3} \mathrm{m}$$. 
Q.5: A kite is flying at a height of 60 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 :
Let K be the kite and the string is tied to point P on the ground.
$$\begin{array}{l}{\text { In } \Delta \mathrm{KLP}_{,}} \\ {\frac{\mathrm{KL}}{\mathrm{KP}}=\sin 60^{\circ}} \\ {\frac{60}{\mathrm{KP}}=\frac{\sqrt{3}}{2}} \\ {\mathrm{KP}=\frac{120}{\sqrt{3}}=40 \sqrt{3} \mathrm{m}} \\ {\text { Hence, the length of the string is } 40 \sqrt{3} \mathrm{m} \text { . }}\end{array}$$ 
Q.6: A 1.5 m tall boy is standing at some distance from a 30 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 :
Let the boy was standing at point S initially. He walked towards the building and reached at point T.
It can be observed that
$$\begin{array}{l}{\mathrm{PR}=\mathrm{PQ}-\mathrm{RQ}} \\ {=(30-1.5) \mathrm{m}=28.5 \mathrm{m}=\frac{57}{2} \mathrm{m}} \\ {\text { In } \Delta \mathrm{PAR}_{,}} \\ {\frac{\mathrm{PR}}{\mathrm{AR}}=\tan 30^{\circ}} \\ {\frac{57}{\mathrm{ARR}}=\frac{1}{\sqrt{3}}} \\ {\mathrm{AR}=\left(\frac{57}{2} \sqrt{3}\right) \mathrm{m}}\end{array}$$
In $$\triangle \mathrm{PRB}$$,
$$\begin{array}{l}{\frac{P R}{B R}=\tan 60^{\circ}} \\ {\frac{57}{2 B R}=\sqrt{3}} \\ {B R=\frac{57}{2 \sqrt{3}}=\left(\frac{19 \sqrt{3}}{2}\right) m} \\ {S T=A B} \\ {=A R-B R=\left(\frac{57 \sqrt{3}}{2}-\frac{19 \sqrt{3}}{2}\right) m} \\ {=\left(\frac{38 \sqrt{3}}{2}\right) m=19 \sqrt{3} \mathrm{m}}\end{array}$$
Hence, he walked $$19 \sqrt{3}$$ m towards the building. 
Q.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 m high building are 45$$^{\circ}$$ and 60$$^{\circ}$$respectively. Find the height of the tower.
Ans :
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 $$\triangle B C D$$,
$$\begin{array}{l}{\frac{\mathrm{BC}}{\mathrm{CD}}=\tan 45^{\circ}} \\ {\frac{20}{\mathrm{CD}}=1} \\ {\mathrm{CD}=20 \mathrm{m}} \\ {\text { In } \Delta \mathrm{ACD}} \\ {\frac{\mathrm{AC}}{\mathrm{CD}}=\tan 60^{\circ}}\end{array}$$
$$\begin{array}{l}{\frac{\mathrm{AB}+\mathrm{BC}}{\mathrm{CD}}=\sqrt{3}} \\ {\frac{\mathrm{AB}+20}{20}=\sqrt{3}} \\ {\mathrm{AB}=(20 \sqrt{3}-20) \mathrm{m}} \\ {=20(\sqrt{3}-1) \mathrm{m}}\end{array}$$
Therefore, the height of the transmission tower is $$20(\sqrt{3}-1)$$m. 
Q.8: A statue, 1.6 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 :
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.
$$\begin{array}{l}{\text { In } \Delta B C D,} \\ {\frac{B C}{C D}=\tan 45^{\circ}} \\ {\frac{B C}{C D}=1} \\ {B C=C D} \\ {\text { In } \triangle A C D_{,}}\end{array}$$
$$\begin{array}{l}{\frac{\mathrm{AB}+\mathrm{BC}}{\mathrm{CD}}=\tan 60^{\circ}} \\ {\frac{\mathrm{AB}+\mathrm{BC}}{\mathrm{BC}}=\sqrt{3}} \\ {1.6+\mathrm{BC}=\mathrm{BC} \sqrt{3}} \\ {\mathrm{BC}(\sqrt{3}-1)=1.6}\end{array}$$
\begin{aligned} \mathrm{BC} &=\frac{(1.6)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\ &=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3})^{2}-(1)^{2}} \\ &=\frac{1.6(\sqrt{3}+1)}{2}=0.8(\sqrt{3}+1) \end{aligned}
Therefore, the height of the pedestal is 0.8 $$(\sqrt{3}+1)$$m. 
Q.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 m high, find the height of the building.
Ans :
Let AB be the building and CD be the tower.
$$\begin{array}{l}{\text { In } \Delta \mathrm{CDB}_{1}} \\ {\frac{\mathrm{CD}}{\mathrm{BD}}=\tan 60^{\circ}} \\ {\frac{50}{\mathrm{BD}}=\sqrt{3}} \\ {\mathrm{BD}=\frac{50}{\sqrt{3}}}\end{array}$$
$$\begin{array}{l}{\text { In } \triangle A B D,} \\ {\frac{A B}{B D}=\tan 30^{\circ}} \\ {A B=\frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}}=\frac{50}{3}=16 \frac{2}{3}}\end{array}$$
Therefore, the height of the building is $$16 \frac{2}{3} m$$. 
Q.10: Two poles of equal heights are standing opposite each other an either side of the
road, which is 80 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 :
Let AB and CD be the poles and O is the point from where the elevation angles are measured.
$$\begin{array}{l}{\text { In } \triangle A B O_{,}} \\ {\frac{A B}{B O}=\tan 60^{\circ}} \\ {\frac{A B}{B O}=\sqrt{3}} \\ {B O=\frac{A B}{\sqrt{3}}}\end{array}$$
$$\begin{array}{l}{\text { In } \Delta \mathrm{CDO} \text { , }} \\ {\frac{\mathrm{CD}}{\mathrm{DO}}=\tan 30^{\circ}} \\ {\frac{\mathrm{CD}}{80-\mathrm{BO}}=\frac{1}{\sqrt{3}}} \\ {\mathrm{CD} \sqrt{3}=80-\mathrm{BO}} \\ {\mathrm{CD} \sqrt{3}=80-\frac{\mathrm{AB}}{\sqrt{3}}} \\ {\mathrm{CD} \sqrt{3}+\frac{\mathrm{AB}}{\sqrt{3}}=80}\end{array}$$
Since the poles are of equal heights,
CD = AB
$$\begin{array}{l}{\mathrm{CD}\left[\sqrt{3}+\frac{1}{\sqrt{3}}\right]=80} \\ {\mathrm{CD}\left(\frac{3+1}{\sqrt{3}}\right)=80} \\ {\mathrm{CD}=20 \sqrt{3} \mathrm{m}} \\ {\mathrm{BO}=\frac{\mathrm{AB}}{\sqrt{3}}=\frac{\mathrm{CD}}{\sqrt{3}}=\left(\frac{20 \sqrt{3}}{\sqrt{3}}\right) \mathrm{m}=20 \mathrm{m}} \\ {\mathrm{DO}=\mathrm{BD}-\mathrm{BO}=(80-20) \mathrm{m}=60 \mathrm{m}}\end{array}$$
Therefore, the height of poles is $$20 \sqrt{3}$$ m and the point is 20 m and 60 m far from these poles. 
Q.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 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 :
Let AB be a building and CD be a cable tower.
$$\begin{array}{l}{\text { In } \triangle A B D,} \\ {\frac{A B}{B D}=\tan 45^{\circ}} \\ {\frac{7}{B D}=1} \\ {B D=7 m} \\ {\text { In } \Delta A C E_{,}}\end{array}$$
\begin{aligned} A C &=B D=7 m \\ \frac{C E}{A E} &=\tan 60^{\circ} \\ \frac{C E}{7} &=\sqrt{3} \\ C E &=7 \sqrt{3} m \\ C D &=C E+E D=(7 \sqrt{3}+7) m \end{aligned}
Therefore, the height of the cable tower is $$7(\sqrt{3}+1) \mathrm{m}$$.
$$\begin{array}{l}{\frac{\mathrm{AB}}{\mathrm{BC}}=\tan 60^{\circ}} \\ {\frac{\mathrm{AB}}{\mathrm{BC}}=\sqrt{3}} \\ {\mathrm{BC}=\frac{\mathrm{AB}}{\sqrt{3}}} \\ {\text { In } \triangle \mathrm{ABD}}\end{array}$$
$$\begin{array}{l}{\frac{A B}{B D}=\tan 30^{\circ}} \\ {\frac{A B}{B C+C D}=\frac{1}{\sqrt{3}}}\end{array}$$
$$\begin{array}{l}{\frac{A B}{\frac{A B}{\sqrt{3}}+20}=\frac{1}{\sqrt{3}}} \\ {\frac{A B \sqrt{3}}{A B+20 \sqrt{3}}=\frac{1}{\sqrt{3}}} \\ {3 A B=A B+20 \sqrt{3}} \\ {2 A B=20 \sqrt{3}} \\ {A B=10 \sqrt{3} \mathrm{m}}\end{array}$$
$$\mathrm{BC}=\frac{\mathrm{AB}}{\sqrt{3}}=\left(\frac{10 \sqrt{3}}{\sqrt{3}}\right) \mathrm{m}=10 \mathrm{m}$$
Therefore, the height of the tower is $$10 \sqrt{3}$$ m and the width of the canal is 10 m. 
Q.12: From the top of a 7 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 :
Let AB be a building and CD be a cable tower.
In ΔABD,
$$\begin{array}{l}{\frac{\mathrm{AB}}{\mathrm{BD}}=\tan 45^{\circ}} \\ {\frac{7}{\mathrm{BD}}=1} \\ {\mathrm{BD}=7 \mathrm{m}}\end{array}$$
In ΔACE,
AE = BD = 7 m
$$\begin{array}{l}{\frac{\mathrm{CE}}{\mathrm{AE}}=\tan 60^{\circ}} \\ {\frac{\mathrm{CE}}{7}=\sqrt{3}} \\ {\mathrm{CE}=7 \sqrt{3} \mathrm{m}} \\ {\mathrm{CD}=\mathrm{CE}+\mathrm{ED}=(7 \sqrt{3}+7) \mathrm{m}}\end{array}$$
$$\begin{array}{c}{=7(\sqrt{3}+1) \mathrm{m}} \\ {\text { Therefore, the height of the cable tower is } 7(\sqrt{3}+1) \mathrm{m}}\end{array}$$ 
Q.13: As observed from the top of a 75 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 :
Let AB be the lighthouse and the two ships be at point C and D respectively.
In triangle ABC ,$$\begin{array}{l}{\frac{\mathrm{AB}}{\mathrm{BC}}=\tan 45^{\circ}} \\ {\frac{75}{\mathrm{BC}}=1} \\ {\mathrm{BC}=75 \mathrm{m}}\end{array}$$
$$\begin{array}{l}{\text { In } \triangle A B D \text { , }} \\ {\frac{A B}{B D}=\tan 30^{\circ}} \\ {\frac{75}{B C+C D}=\frac{1}{\sqrt{3}}} \\ {\frac{75}{75+C D}=\frac{1}{\sqrt{3}}} \\ {75 \sqrt{3}=75+C D} \\ {75(\sqrt{3}-1) m=C D}\end{array}$$
Therefore, the distance between the ships is  $$75(\sqrt{3}-1)$$ m. 
Q.14: A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 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 the angle of elevation reduces to 30$$^{\circ}$$

Ans :
Let the initial position A of balloon change to B after some time and CD be the girl.In $$\triangle$$ ACE,
$$\begin{array}{l}{\frac{\mathrm{AE}}{\mathrm{CE}}=\tan 60^{\circ}} \\ {\frac{\mathrm{AF}-\mathrm{EF}}{\mathrm{CE}}=\tan 60^{\circ}} \\ {\frac{88.2-1.2}{\mathrm{CE}}=\sqrt{3}} \\ {\frac{87}{\mathrm{CE}}=\sqrt{3}} \\ {\mathrm{CE}=\frac{\mathrm{g} 7}{\sqrt{3}}=29 \sqrt{3} \mathrm{m}}\end{array}$$
$$\begin{array}{l}{\text { In } \Delta B C G,} \\ {\frac{B G}{C G}=\tan 30^{\circ}} \\ {\frac{88.2-1.2}{C G}=\frac{1}{\sqrt{3}}} \\ {\text { Distance travelled by balloon }=E G=C G-C E} \\ {=(87 \sqrt{3}-29 \sqrt{3}) \mathrm{m}} \\ {=58 \sqrt{3} \mathrm{m}}\end{array}$$ 
Q.15: A straight highway leads to the foot of a tower. A man standing at the top of the
tower observes a car as 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.
Ans :
$$\begin{array}{l}{\text { Let } A B \text { be the tower. }} \\ {\text { Initial position of the car is C, which changes to D after six seconds. }} \\ {\text { In } \triangle A D B \text { , }} \\ {\frac{A B}{D B}=\tan 60^{\circ}} \\ {\frac{A B}{D B}=\sqrt{3}} \\ {D B=\frac{A B}{\sqrt{3}}}\end{array}$$
$$\begin{array}{l}{\text { In } \triangle \mathrm{ABC}} \\ {\frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ}} \\ {\frac{\mathrm{AB}}{\mathrm{BD}+\mathrm{DC}}=\frac{1}{\sqrt{3}}} \\ {\mathrm{AB} \sqrt{3}=\mathrm{BD}+\mathrm{DC}} \\ {\mathrm{AB} \sqrt{3}=\frac{\mathrm{AB}}{\sqrt{3}}+\mathrm{DC}}\end{array}$$
$$\mathrm{DC} = \mathrm{AB} \sqrt{3}-\frac{\mathrm{AB}}{\sqrt{3}}=\mathrm{AB}\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) \\ =\frac{2 \mathrm{AB}}{\sqrt{3}} \\ \text { Time taken by the car to travel distance DC } \text { (i.e., } \frac{2 \mathrm{AB}}{\sqrt{3}} )=6 \text { seconds }$$
$$\begin{array}{l}{\text { Time taken by the car to travel distance DB }\left(\text { i.e., } \frac{A B}{\sqrt{3}}\right)=\frac{6}{\frac{2 A B}{\sqrt{3}}} \times \frac{A B}{\sqrt{3}}} \\ {=\frac{6}{2}=3 \text { seconds }}\end{array}$$ 
Q.16: The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 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 m .
Ans :
Let AQ be the tower and R, S are the points 4 m, 9 m away from the base of the tower respectively.
The angles are complementary. Therefore, if one angle is $$\theta$$, the other will be $$90-\theta.$$
$$\begin{array}{l}{\text { In } \triangle A Q R,} \\ {\frac{A Q}{Q R}=\tan \theta} \\ {\frac{A Q}{4}=\tan \theta} \\ {\text { In } \Delta A Q S,}\end{array}$$
$$\begin{array}{l}{\frac{\mathrm{AQ}}{\mathrm{SQ}}=\tan (90-\theta)} \\ {\frac{\mathrm{AQ}}{9}=\cot \theta} \\ {\text { On multiplying equations ( } i ) \text { and (ii), we obtain }} \\ {\left(\frac{\mathrm{AQ}}{4}\right)\left(\frac{\mathrm{AQ}}{9}\right)=(\tan \theta) \cdot(\cot \theta)}\end{array}$$
$$\begin{array}{l}{\frac{\mathrm{AQ}^{2}}{36}=1} \\ {\mathrm{AQ}^{2}=36} \\ {\mathrm{AQ}=\sqrt{36}=\pm 6}\end{array}$$
However, height cannot be negative.
Therefore, the height of the tower is 6 m. 

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