NCERT Solutions for Class 10 Maths Chapter 9 Exercise 9.1

Mathematics is an important subject for students in all facets of their lives, not just in school. Therefore, having a solid understanding of Mathematics will help them not just in school, but also in their daily lives. Class 10 Mathematics is extremely important since students study various important topics with many applications. By properly learning Mathematics and gaining high grades on board exams, students can lay a solid foundation for further studies. One of the most critical and essential subjects for students in Class 10 is Mathematics. It is sometimes regarded as the most difficult subject among students in Class 10. It calls for extensive practise with the problems as well as a deep understanding of the principles. Typically, students learn and retain every concept in the curriculum, but they do not practise answering questions enough. As a result, they are put at a disadvantage when sitting in the board exams. Therefore, it is crucial that they practise the questions in the NCERT textbook exercises. As these books are suggested by the CBSE, NCERT books serve as the primary source of instruction in the majority of schools. This increases the need for NCERT solutions. For senior classes, such as Class 10, NCERT solutions are crucial for board exam preparation. To give students plenty of practise before the board exam, Extramarks offers NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 and other study modules.

During a student’s education, Class 10 is without a doubt one of the most crucial years. Students must appear for the board exams this year, which have a significant impact on their academic careers. As the board exams are a new experience for them, students in Class 10 face problems that they have never encountered before. All throughout a person’s life, the knowledge acquired in Class 10 is applied in a variety of ways. Every topic covered in Class 10 has the potential to help students succeed. Students in Class 10 compete against peers from throughout the nation in addition to their classmates, unlike in previous classes. Students frequently become worried when competing against lakhs of other students. Sometimes, students become mentally nervous about taking board exams. They consequently perform poorly in the exams. In order to overcome psychological difficulties, students must develop certain effective tactics. Students in Class 10 must put in a lot of work to succeed in the Class 10 board exams. Their careers may benefit from performing well on the Class 10 board exams. The results of the Class 10 board exams play a significant role in determining admissions to various higher education institutions. The results of Class 10 are taken into account for admission to colleges and universities as well as for securing work opportunities in the future because they are a crucial indicator of a student’s academic performance. Students’ confidence will increase as a result of doing well in the Class 10 board exams, and they will not be afraid of future exams. As a result, they will be able to do better in future exams. In Class 10, students have to study a number of subjects. Therefore, Class 10 serves as a year when students make crucial choices for their future professional careers. After Class 10, students have to decide which path to pursue based on their interests and skills. They can determine their areas of interest and then decide on a stream and subject that is right for them in the future.

Extramarks provides easy access to the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1. Finding NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 is frequently very difficult for students in Class 10. There are numerous NCERT solutions available online. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 produced by qualified teachers are available on several websites. Students must, however, independently confirm the reliability of these sources. Online learning platforms are not always trustworthy. Students may benefit from a select few websites that provide thoroughly written NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1. Numerous platforms also do not offer thorough step-by-step NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1. This has a detrimental effect on the students. On the other hand, the Extramarks website has reliable NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1. The Extramarks website, which offers NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1, makes it easier for students in Class 10 to obtain precise solutions developed by experts.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry (Ex 9.1) H2 – Exercise 9.1

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Access NCERT Solutions for Class 10 Maths Chapter 9 – Trigonometry

Prior to the board exams, students must have access to a large number of questions as well as well-explained solutions. The vast majority of question banks mainly depend on the NCERT textbook questions. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 make answering questions easier. When students understand and learn quickly, they are much more confident in any examination.Students can also devote more time to all other elements of their studies whilst also managing to keep their minds cool. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 completely answer all of the questions in Ex 9.1 Class 10. As a result, repeatedly practising the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 will enable students to answer such questions quickly and accurately in exams.

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

The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 are an important resource for Class 10 students who want to get a head start on their preparation for the Class 10 board examinations in Mathematics. The CBSE recommends the NCERT-published Mathematics textbook for Class 10. Several other boards include the NCERT textbook in their curricula as well. Chapter 9 titled – Some Applications of Trigonometry is introduced to Class 10 students of the NCERT Mathematics book. Trigonometry is applied to find the heights and distances of various objects without actually measuring them. Trigonometry is used by astronomers to find the distances of planets and stars from the Earth. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1, along with the study material provided by Extramarks, cover all of the important concepts covered by the CBSE syllabus for the board examinations. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 are developed by expert teachers to prepare students to effectively solve questions that appear in the CBSE board examinations. Practising the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 also facilitates students’ preparation for various competitive examinations like the Olympiad and others. To be thorough with the problems, students should practise the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 at least twice.

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The Central Board of Secondary Education (CBSE) is a national school education board in India. It is the board that has the most students enrolled in the country. Furthermore, it affiliates with thousands of schools. The CBSE is also in charge of developing the Class 10 curriculum. The NCERT textbooks are prescribed by the board for the students in Class 10. The NCERT textbook includes a significant number of exam-relevant questions. These questions assist students in becoming thoroughly acquainted with the topics, which benefits them not only in the examinations but also in their subsequent studies. Students must understand the concepts in the textbook in order to succeed in their higher education as well. However, many Class 10 students find it difficult to study for their exams. There are many new concepts in the Class 10 Mathematics curriculum. In order to do well on the board exams, students must have access to every required study material. The NCERT textbook represents one of the most crucial study tools for preparing for board exams. The exercises include a variety of questions. Prior to taking the board exams, students must correctly answer these questions. They face a number of difficulties as they attempt to answer these questions. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1, which are available on Extramarks, can assist students in this area.

Students preparing for the Class 10 board exams must put in a lot of effort. Students preparing for board exams must not only work hard but also make informed choices. If students want to achieve extraordinary results from their preparation, they must develop a strong approach. They must create a study plan for each subject and make every effort to stick to it. They must have everything planned out from the beginning. First, they should go over the course syllabus, and then thoroughly study each topic in accordance with it. Students should practise textbook questions while also writing notes on important points. They should solve the examples in the NCERT textbook and learn the methods of solving each type of question. They should also try to answer all of the questions in the exercises on their own. Extramarks’ NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 can help them in doing so. Formulas used to answer the questions should also be noted. They can also summarise the steps they took to answer the questions. Each topic should then be revised, with a greater emphasis placed on the areas where they lag. Finally, they should take a mock test to ensure that they understand the material completely. They can also solve sample question papers as well as past years’ papers that are fully accessible on Extramarks.

Students should take several simple steps to answer the questions in this exercise quickly and efficiently. Mathematics requires students to write step-by-step answers.  If they are to receive the highest possible marks, they must be aware of the procedures to follow when writing answers.Students should read and comprehend the topic explanations in the textbook. Furthermore, if they are having difficulty grasping the topics, they can rely on the study materials provided by Extramarks. Once they have a clear understanding of the concepts, they must read through the examples in the NCERT book. As a result, they will gain an understanding of how to approach the questions. After that, they should try to answer each question in the exercise in order. They should try to solve the problems on their own before reviewing the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1. Students should try their best to recall every method that could be used to answer the questions. They should check their answers after answering each question using the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1. They can use the solutions to locate their mistakes. Students should try to solve such problems repeatedly until they can do so appropriately. They should also check the solutions to get answers to the questions they have difficulty answering. Students should try to answer the exercise questions until they have perfected the correct answers. They will get more accustomed to dealing with the questions and will be able to write faster as a result. Students should try to complete the exercise as quickly as possible. This would allow them to complete the examination paper on time by solving such questions more easily and effectively.

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NCERT Solutions for Class 10 Maths PDFs

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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° (see the following figure).


Ans.

In the given figure above, we have        sin 30°= AB AC = AB 20 or 1 2 = AB 20 or AB= 20 2 =10 Therefore, height of the pole is 10 m. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A550@

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° 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 the tree AB bends at point C. Foot of the tree is at A and CB’ is the broken part of the tree i.e, BC = CB’.

Now, in the above figure, we have       tan 30°=ACAB’=AC8or 13=AC8or AC=83Also, cos 30°=AB’CB’=8CB’or 32=8CB’or CB’=163Therefore, height of the tree=AC+CB’=83+163=243=83  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° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Ans.
As per given information, we have the following figure.

In ΔABC, we have       sin 30°=ABAC=1.5ACor 12=1.5ACor AC=1.5×2=3 mTherefore, length of the slide for the children belowthe age of 5 years is 3 m.

Also, for elder children, we have the following slide.

In ΔPQR, we have       sin 60°=PQPR=3PRor 32=3PRor PR=3×23=23 mTherefore, length of the slide for the elder children is 23 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°. Find the height of the tower.

Ans.
Let PQ is a tower with foot at Q and R is a point on the ground such that distance between Q and R is 30 m.

In ΔPQR, we have       tan 30°=PQQR=PQ30or 13=PQ30or PQ=303=103 mTherefore, height of the tower is 103 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°. Find the length of the string, assuming that there is no slack in the string.

Ans.
Let the kite is flying at point A and AC is the string of the kite tied at point C on the ground.

In ΔABC, we have       sin 60°=ABAC=60ACor 32=60ACor AC=60×23=403 mTherefore, length of the string is 403 m.

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° to 60° as he walks towards the building. Find the distance he walked towards the building.

Ans.
Let AB be a building of height 30 m. Let the boy is initially at D and then from D moves to point E. We have to find the distance DE

In ΔACE, we have       tan 60°=ACEC=28.5ECor 3=28.5ECor EC=28.53 mNow, In ΔACD, we have       tan 30°=ACCD=28.5EC+DEor 13=28.5EC+DEor EC+DE=28.53 or DE=28.53 −EC=28.53−28.53or DE=28.5×3−28.53=573=19×33=193 mTherefore, the required distance is 193 m.

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° and 60° respectively. Find the height of the tower.

Ans.
Let BC be a building on which a transmission tower AB is fixed. Let D be a point on the ground.

In ΔBCD, we have       tan 45°=BCDC=20DCor 1=20DCor DC=20 mNow, in Δ ACD, we havetan 60°=ACDC=AC20or 3=AC20or AC=203 or AC=AB+BC=203 or AB=203−BC=203−20=20(3−1)Therefore, height of the tower is 20(3−1) m.

Q.8 A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Ans.
Let AB be the statue on the pedestal BC. Let D be a point on the ground.

In ΔBCD, we have       tan 45°=BCDCor 1=BCDCor DC=BCNow, in Δ ACD, we havetan 60°=ACDC=AB+BCBCor 3=1.6+BCBCor 3 BC=1.6+BCor BC(3−1) =1.6or BC=1.6(3−1)=1.6(3−1)×3+13+1=0.8(3+1) mTherefore, height of the pedestal is 0.8(3+1) m.

Q.9 The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Ans.
Let AB be the height of the building and CD be the height of the tower.

In ΔBCD, we have       tan 60°=CDBDor 3=50BDor BD=503Now, in Δ ABC, we havetan 30°=ABBDor 13=AB503or AB=13×503=503=1623Therefore, height of the building is 1623 m.

Q.10 Two poles of equal heights are standing opposite each other on 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° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Ans.
Let AB and CD be the two poles of equal height. Let O be a point on the road from where elevation angles are measured.

In Δ ABO, we have       tan 60°=ABBOor 3=ABBOor BO=AB3Now, in Δ CDO, we havetan 30°=CDODor 13=CDOD=CDBD−BO=CD80−BOor CD3=80−BO=80−AB3or AB3=80−AB3                             [AB=CD as heights of the poles are equal]or AB3+AB3   =80or 4AB3=80or AB=203Therefore, height of each pole is 203 m. BO=AB3=2033=20 mOD=BD−BO=80−20=60 mTherefore, the required distances are 20 m and 60 m.

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°. 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° (see the following figure). Find the height of the tower and the width of the canal.


Ans.
We have the following figure.

In ΔABD, we havetan 30° = ABDB = ABBC+CDor 13 = ABBC+20or BC+20 = AB3or BC = AB3−20                                                                  ...(1)Now, In Δ ABC, we have       tan 60° = ABBCor 3 = ABBCor AB = BC3                                                                            ...(2)From equations (1) and (2), we have         BC = AB3−20or BC = BC3 ×3−20=3BC-20or 3BC−BC = 20or 2BC = 20or BC = 10 mOn putting this value of BC in equation (2), we get AB = 103 mTherefore, height of the tower is 103 m and width ofthe 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° and the angle of depression of its foot is 45°. Determine the height of the tower.

Ans.
Let AB is a cable tower and CD is a building.

We have, CD = BE = 7 m, DB = CEIn Δ CDB, we have     tan 45° = CDDBor 1 = 7DBor DB = 7 mNow, In Δ ACE, we have       tan 60° = AECEor 3 = AECEor AE = CE3  = DB3 = 73 m        [CE = DB]Therefore, height of the tower = AB = AE+BE = 73+7 = 7(1+3) m

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° and 45°. 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 a lighthouse and ships be at points C and D.It is given that AB = 75 m. We have to find the distance CD.In Δ ABC, we have     tan 45° = ABBCor 1 = ABBCor BC = AB = 75 mNow, In Δ ABD, we have       tan 30° = ABDBor 13 = 75DBor DB = 753or DB = BC+CD = 753or CD = 753−BC = 753−75 = 75 (3−1)Therefore, distance between the two ships = 75 (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°. After some time, the angle of elevation reduces to 30° (see the following figure). Find the distance travelled by the balloon during the interval.


Ans.

Let the balloon be initially at A and then it moves to B.Let CD be the girl.In Δ ACE, we have        tan 60° = AECEor 3 = AECEor CE = AE3 = AF−EF3 = 88.2−1.23 = 873 = 293In Δ BCG, we have        tan 30° = BGCGor 13 = 88.2−1.2CGor CG = 873 mTherefore,distance travelled by the balloon = EG                                         = CG-CE                                         = 873−293                                         = 583 m

Q.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°, 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°. Find the time taken by the car to reach the foot of the tower from this point.

Ans.

Let AB be a man and BC be a tower. Let the car is initiallyat D and after 6 seconds it moves to point E.In Δ ACE, we have     tan 60° = ACCEor 3 = ACCEor AC = 3 CE ...(1)Now, In Δ ACD, we have       tan 30° = ACCDor 13 = ACCDor CD = AC3or CD = 3×3 CE = 3CE from (1), AC = 3 CE or DE+CE = 3CEor DE = 2CEor CE = DE2The car takes 6 seconds to cover the distance DE.Therefore, it will take 3 seconds to cover the half of thedistance DE i.e., the car will take 3 seconds to reach thefoot of the tower from the point E.

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 AB is a tower. Let C and D respectively are two points at a distance of 4 m and 9 m from the base of the tower.

Let angles of elevation of the top of the tower from D be θ and from C be 90°−θ.In Δ ABC, we have     tan (90°−θ) = ABBCor cotθ = AB4or AB = 4cotθ ...(1)Now, In Δ ABD, we have       tan θ = ABBDor  tan θ = AB9or   AB = 9 tanθ ...(2)From (1) and (2), we have 9 tanθ = 4cotθor   tanθcotθ = 49or   tan2θ = 49or   tanθ = 23On putting this value of tanθ in (2), we getAB = 9 tanθ = 9×23 = 6 mTherefore, the height of the tower is 6 m.

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1. What is the number of questions in the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1?

The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 include answers to 16 questions from the exercise.

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NCERT solutions are essential for secondary classes such as Class 10 in order to prepare for board exams. Studying modules such as the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 allow students to practice extensively before taking any examination or test. Answers are insufficient because students must understand the steps taken to arrive at the answer. The clarity of the process and steps can help students improve their ability to answer any question that arises in the board exam.

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