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Chapter 6: Trigonometry — Online MCQ Test

MATHS · CLASS 10th · Tamil Nadu State Board

Practice Chapter 6: Trigonometry with a free chapter-wise online MCQ test for Tamil Nadu State Board CLASS 10th MATHS. This chapter covers: Trigonometric ratios Trigonometric identities Sine Cosine Tangent Heights and distances Angle of elevation Angle of depression Complementary angles Applications of trigonometry. AI-generated questions from basic to board-exam level, with instant results and explanations.

10
Questions
20m
Time Limit
3
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  • 10 random questions from this chapter (mixed difficulty)
  • Questions you've seen before won't repeat until the pool resets
  • You have 20 minutes — exam auto-submits when time is up
  • Maximum 3 attempts per chapter
  • Results and explanations shown immediately after submission
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Chapter 6: Trigonometry — Important Questions & Answers (FAQ)

Frequently asked questions from Tamil Nadu State Board CLASS 10th MATHS — Chapter 6: Trigonometry, with answers and explanations. These are sample questions; the exam has its own separate question set.

In a right-angled triangle, sin θ is equal to ______.
  • A. opposite side / hypotenuse ✓
  • B. adjacent side / hypotenuse
  • C. opposite side / adjacent side
  • D. hypotenuse / opposite side
Answer: A. opposite side / hypotenuse
The sine ratio is defined as the length of the side opposite to θ divided by the hypotenuse.
Which of the following is a fundamental trigonometric identity?
  • A. sin A + cos A = 1
  • B. sin² A - cos² A = 1
  • C. sin² A + cos² A = 1 ✓
  • D. tan² A + cot² A = 1
Answer: C. sin² A + cos² A = 1
The basic identity connecting sine and cosine is sin² A + cos² A = 1.
If sin A = 3/5 and A is an acute angle, then cos A is
  • A. 5/3
  • B. 3/4
  • C. 4/5 ✓
  • D. 5/4
Answer: C. 4/5
Using sin² A + cos² A = 1, cos² A = 1 - 9/25 = 16/25. Since A is acute, cos A = 4/5.
From the top of a tower, the angle of depression of a car is 30°. The car moves 60 m towards the tower and the angle of depression becomes 60°. The height of the tower is
  • A. 30 m
  • B. 30√3 m ✓
  • C. 60√3 m
  • D. 90 m
Answer: B. 30√3 m
Let the height be h. If the initial distance is x, then h/x = 1/√3 and h/(x - 60) = √3. Solving gives h = 30√3 m.
From the top of a lighthouse 100 m high, the angles of depression of two ships on the same side of it are 45° and 30°. The distance between the ships is
  • A. 100(√3 - 1) m ✓
  • B. 100(√3 + 1) m
  • C. 100√3 m
  • D. 200 m
Answer: A. 100(√3 - 1) m
The horizontal distances are 100/tan 45° = 100 m and 100/tan 30° = 100√3 m. Their difference is 100(√3 - 1) m.

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