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Rajasthan Board Class 10 (RBSE) Introduction to Trigonometry — practice questions

Explore 13 free Introduction to Trigonometry questions in Rajasthan Board Class 10 (RBSE) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Introduction to Trigonometry questions and exercises with answers

Showing questions 1–13 of 13.

In a right triangle, for an acute angle θ, sin θ is defined as:Read the answer and worked solution →In a right triangle, for an acute angle θ, cos θ is defined as:Read the answer and worked solution →In terms of sine and cosine, tan θ equals:Read the answer and worked solution →The fundamental trigonometric identity is:Read the answer and worked solution →The value of sin 30° is:Read the answer and worked solution →The value of cos 60° is:Read the answer and worked solution →The value of tan 45° is:Read the answer and worked solution →The value of sin 90° is:Read the answer and worked solution →The value of cos 0° is:Read the answer and worked solution →The identity $1 + \tan^2\theta$ equals:Read the answer and worked solution →The expression $\sin(90^\circ - \theta)$ is equal to:Read the answer and worked solution →The value of tan 0° is:Read the answer and worked solution →The reciprocal of sin θ is:Read the answer and worked solution →

Worked practice questions

In a right triangle, for an acute angle θ, sin θ is defined as:

  1. hypotenuse / opposite side
  2. opposite side / hypotenuse
  3. opposite side / adjacent side
  4. adjacent side / hypotenuse
Show answer and solution

Correct answer: B. opposite side / hypotenuse

sin θ = side opposite θ / hypotenuse.

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In a right triangle, for an acute angle θ, cos θ is defined as:

  1. hypotenuse / adjacent side
  2. adjacent side / hypotenuse
  3. opposite side / hypotenuse
  4. adjacent side / opposite side
Show answer and solution

Correct answer: B. adjacent side / hypotenuse

cos θ = side adjacent θ / hypotenuse.

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In terms of sine and cosine, tan θ equals:

  1. sin θ / cos θ
  2. cos θ / sin θ
  3. 1 / sin θ
  4. sin θ × cos θ
Show answer and solution

Correct answer: A. sin θ / cos θ

tan θ = (opposite/hyp)/(adjacent/hyp) = sin θ / cos θ.

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