CBSE Class 10 Mathematics Applications Of Trigonometry Worksheet Set C

Read and download free pdf of CBSE Class 10 Mathematics Applications Of Trigonometry Worksheet Set C. Students and teachers of Class 10 Mathematics can get free printable Worksheets for Class 10 Mathematics Chapter 9 Applications of Trigonometry in PDF format prepared as per the latest syllabus and examination pattern in your schools. Class 10 students should practice questions and answers given here for Mathematics in Class 10 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 10 Mathematics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems with solutions on daily basis to get more score in school exams and tests

Worksheet for Class 10 Mathematics Chapter 9 Applications of Trigonometry

Class 10 Mathematics students should refer to the following printable worksheet in Pdf for Chapter 9 Applications of Trigonometry in Class 10. This test paper with questions and answers for Class 10 will be very useful for exams and help you to score good marks

Class 10 Mathematics Worksheet for Chapter 9 Applications of Trigonometry

 

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

 

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CBSE Class 10 Mathematics Chapter 9 Applications of Trigonometry Worksheet

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