NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1

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Jasmine Grover

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1 is provided in this article. Class 10 Maths Chapter 8 Introduction to Trigonometry covers important concepts like trigonometric ratios, trigonometry table, trigonometric identities and formulas. Chapter 8 exercise 8.1 includes questions based on trigonometric ratios.

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CBSE X Related Questions

  • 1.
    Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.


      • 2.
        A kite is flying at a height of \(60 \text{ m}\) above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as \(30^{\circ}\). From the bottom of the same building, the angle of elevation of kite is \(45^{\circ}\). Find the length of the string and height of roof from the ground. (Use \(\sqrt{3} = 1.73\))


          • 3.
            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.

            • 4.
              The natural number 1 is :

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

              • 5.
                Prove that :
                \(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).


                  • 6.
                    \(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

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

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