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

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The NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry are provided in this article. The chapter discusses how trigonometry is helpful in finding the height and distance of various objects without measuring them directly. Trigonometry is mostly used in geography and navigation to locate the position of a place in relation to latitude and longitude.

Class 10 Maths Chapter 9 Some Applications of Trigonometry belongs to Unit 5 Trigonometry which has a weightage of 12 marks in the CBSE Class 10 Maths Examination. The questions from this chapter are based on heights and distances and angles of depression.

Download PDF: NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 


NCERT Solutions for Class 10 Maths Chapter 9

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Important Topics in Class 10 Maths Chapter 9

  • Line of sight and horizontal level

Line of sight is the line from the eye of the observer to the point of the object viewed by the observer.

Horizontal level is a horizontal line through the eye of the observer.

  • The angle of elevation is the angle formed by the line of sight with the horizontal level.
Angle of elevation is relevant for objects that are above horizontal level.
  • The angle of depression is the angle that is formed between the line of sight and the horizontal level.
Angle of depression is is relevant for objects that are below horizontal level.
  • Calculating Heights and Distances: To calculate the heights and distances, we make use of trigonometric ratios. 

Step 1: Draw a line diagram that corresponds to the given problem.

Step 2: Mark all the known angles, heights, and distances. Denote the unknown lengths by variables such as x,y, etc.

Step 3: Use the values of different trigonometric ratios of the angles to find the unknown lengths from the given lengths.


Some Applications of Trigonometry – Related Topics:

CBSE Class 10 Mathematics Study Guides:

CBSE X Related Questions

  • 1.
    If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

      • $x^2 + 5x - 4$
      • $(x + 3) (-x + 8)$
      • $a(x^2 + 5x - 24)$
      • $x^2 - 24$

    • 2.
      The natural number 1 is :

        • a prime number.
        • a composite number.
        • prime as well as composite.
        • neither prime nor composite.

      • 3.
        The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

          • $q - 9p$
          • $p - 9q$
          • $p + 9q$
          • $2p + 9q$

        • 4.
          Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
          Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

            • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
            • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
            • Assertion (A) is true, but Reason (R) is false.
            • Assertion (A) is false, but Reason (R) is true.

          • 5.
            \(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

              • \(\frac{21}{4} \text{ cm}\)
              • \(\frac{28}{3} \text{ cm}\)
              • \(\frac{12}{7} \text{ cm}\)
              • \(5.5 \text{ cm}\)

            • 6.
              Prove that :
              \(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

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