CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set C

Read and download the CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set C in PDF format. We have provided exhaustive and printable Class 10 Mathematics worksheets for Chapter 9 Some Applications of Trigonometry, designed by expert teachers. These resources align with the 2025-26 syllabus and examination patterns issued by NCERT, CBSE, and KVS, helping students master all important chapter topics.

Chapter-wise Worksheet for Class 10 Mathematics Chapter 9 Some Applications of Trigonometry

Students of Class 10 should use this Mathematics practice paper to check their understanding of Chapter 9 Some Applications of Trigonometry as it includes essential problems and detailed solutions. Regular self-testing with these will help you achieve higher marks in your school tests and final examinations.

Class 10 Mathematics Chapter 9 Some Applications of Trigonometry Worksheet with Answers

 

Applications Of Trigonometry

Q.- From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower.
 
Sol. Let AB be the cliff and CD be the tower.
some applications of trigonometry notes 7
Then, AB = 25 m. From B draw BE ⊥ CD.
Let ∠ EBD = ∠ ACB = α.
Now, DE  = tan α and  AB   = tan α
         BE                       AC
  ∴      DE   =  AB   So, DE = AB
           BE       AC                                 [BE = AC]
∴ CD = CE + DE = AB + AB = 2AB = 50m
 
Q.-  The altitude of the sun at any instant is 60º.The height of the vertical pole that will cast a shadow of 30 m is
(A) 30√3 m
(B) 15 m
(C) 30/√3 m 
(D) 15√2 m
 
Sol. Let AB be the pole and AC be its shadow.
Then, θ = 60º and AC = 30 m.
some applications of trigonometry notes 8
AB  = tan 60º 
   AC
=> AB    = √3
      30
=> AB = 30√3 m

 

Q.-  When the sun is 30º above the horizontal, the length of shadow cast by a building 50m high is-
(A) 50√3m
(B) 50√3m
(C) 25 m
(D) 25√3 m
 
Sol. Let AB be the building and AC be its shadow.
Then, AB = 50 m and θ = 30º.
some applications of trigonometry notes 9
∴  AC     = cot 30º = √3
    AB
=> AC   =√3
      50
=> AC = 50√3 cm. 
 
Q.- If the elevation of the sun changed from 30º to 60º, then the difference between the lengths of shadows of a pole 15 m high, made at these two positions, is–
(A) 7.5 m
(B) 15 m
(C) 10√3 m
(D)15/√3 m
 
Sol. When AB = 15m, θ= 30º, then   AC   = tan30º
                                                        AB
 
=> AC = 15/√3 m
 
When AB = 15m, θ = 60º, then AC  = tan60º
                                                AB
=> AC = 15√3m
∴ Diff. in lengths of shadows = ( 15√3 -  15/√3 )
 
=> 30/√3 = 10√3  m
 
Q.- The heights of two poles are 80 m and 62.5 m.If the line joining their tops makes an angle of 45º with the horizontal, then the distance between the poles, is -
(A) 17.5 m
(B) 56.4 m
(C) 12.33 m
(D) 44 m
 
Sol. Let AB and CD be the poles such that
AB = 80 m and CD = 62.5 m.
some applications of trigonometry notes 10
Draw DE ⊥ AB. Then,
∠ EDB = 45º
Now, BE = AB – AE = AB – CD = 17.5
DE   = cot 45º = 1
BE
 
=> DE = BE = 17.5 m.
 
Q.- A tower is 100√3 metres high. Find the angle of elevation of its top from a point 100 metres away from its foot.
 
Sol. Let AB be the tower of height 100√3 metres,and let C be a point at a distance of 100 metres from the foot of the tower.
Let θ be the angle of elevation of the top of the tower from point C.
some applications of trigonometry notes 11
 In ΔCAB, we have
tanθ = AB
           AC
=> tanθ  = 100√3
                  100
= √3
=> = 60º
Hence, the angle of elevation of the top of the tower from a point 100 metres away from its foot is 60º.
 
Q.- A boat is being rowed away from a cliff 150m high. At the top of the cliff the angle of depression of the boat changes from 60º to 45º in 2 minutes. The speed of the boat is –
(A) 2 km/hr
(B) 1.9 km/hr
(C) 2.4 km/hr
(D) 3 km/hr
 
Sol. Let AB be the cliff and C and D be the two positions of the ship. Then, AB = 150 m,
∠ACB = 60º and ∠ADB = 45º.
some applications of trigonometry notes 12
 
Now, AD   = cot 45º = 1
        AB
 
=>  AD    = 1 
       150
=> AD = 150 m.
AC   = cot 60º = 1/√3
AB
=> AC  = 1/√3
     150
=>  AC = 150  = 50√3 = 86.6 m.
               √3
∴ CD = AD – AC = (150 – 86.6) m = 63.4 m
Thus, distance covered in 2 min. = 63.4 m
∴ Speed of the boat
 
(63.4 ×  60) km/ hr.
  2       1000
 
= 1.9 km/hr

 

Please click the link below to download CBSE Class 10 Mathematics Applications Of Trigonometry Worksheet Set C

CBSE Mathematics Class 10 Chapter 9 Some Applications of Trigonometry Worksheet

Students can use the practice questions and answers provided above for Chapter 9 Some Applications of Trigonometry to prepare for their upcoming school tests. This resource is designed by expert teachers as per the latest 2026 syllabus released by CBSE for Class 10. We suggest that Class 10 students solve these questions daily for a strong foundation in Mathematics.

Chapter 9 Some Applications of Trigonometry Solutions & NCERT Alignment

Our expert teachers have referred to the latest NCERT book for Class 10 Mathematics to create these exercises. After solving the questions you should compare your answers with our detailed solutions as they have been designed by expert teachers. You will understand the correct way to write answers for the CBSE exams. You can also see above MCQ questions for Mathematics to cover every important topic in the chapter.

Class 10 Exam Preparation Strategy

Regular practice of this Class 10 Mathematics study material helps you to be familiar with the most regularly asked exam topics. If you find any topic in Chapter 9 Some Applications of Trigonometry difficult then you can refer to our NCERT solutions for Class 10 Mathematics. All revision sheets and printable assignments on studiestoday.com are free and updated to help students get better scores in their school examinations.

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