Class 10 Maths Chapter 8: Introduction to Trigonometry worksheet with answer key
A 20-mark worksheet on Introduction to Trigonometry for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (b) 1517
For angle R the adjacent side is QR = 15 cm and the hypotenuse is PR = 17 cm, so cos R = 15/17. (8/17 is cos P.)
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Answer: (d) 32
cosec A is the reciprocal of sin A, so cosec A = 1 ÷ (2/3) = 3/2.
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Answer: (a) 1213
The side opposite A is BC = 12 cm and the hypotenuse is AC = 13 cm, so sin A = 12/13.
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Answer: (c) 2
cos 0° = 1, sin 90° = 1 and sin 0° = 0, so the sum is 2.
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Answer: (a) 32
tan 45° = 1 and sin 45° cos 45° = (1/√2)(1/√2) = 1/2, so the sum is 3/2.
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tan 45° = 1 and sin 60° = √3/2. The product is (1 + √3/2)(1 − √3/2) = 1 − 3/4 = 1/4.
- tan 45° = 1 and sin 60° = √3/2 (1)
- value = 1/4 (1)
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sin A = 1/3. Let the opposite side be k and the hypotenuse 3k. The adjacent side = √(9k² − k²) = 2√2 k. So cos A = 2√2/3 (positive, as A is acute) and cot A = adjacent/opposite = 2√2 k/k = 2√2.
- finds the third side 2√2 k by Pythagoras (or uses cos A = √(1 − sin²A)) (1)
- cos A = 2√2/3 and cot A = 2√2 (1)
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sin 0° = 0 and cos 90° = 0. tan 0° = sin 0°/cos 0° = 0/1 = 0. tan 90° = sin 90°/cos 90° = 1/0, and division by zero is not defined, so tan 90° is not defined.
- sin 0° = 0 and cos 90° = 0 (1)
- tan 90° not defined, because cos 90° = 0 makes the quotient 1/0 (1)
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Let PQ = 20k and QR = 21k. By Pythagoras, (20k)² + (21k)² = 58², so 400k² + 441k² = 841k² = 3364, k² = 4 and k = 2. So PQ = 40 cm and QR = 42 cm. sin P = QR/PR = 42/58 = 21/29, cos P = PQ/PR = 40/58 = 20/29, tan P = QR/PQ = 42/40 = 21/20.
- sets up 841k² = 3364 (1)
- k = 2, PQ = 40 cm, QR = 42 cm (1)
- sin P = 21/29, cos P = 20/29, tan P = 21/20 (1)
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LHS = sec²A (sec²A − 1). By 1 + tan²A = sec²A, sec²A − 1 = tan²A, so LHS = (1 + tan²A) tan²A = tan²A + tan⁴A = RHS.
- takes sec²A common (1)
- uses sec²A − 1 = tan²A (1)
- writes sec²A = 1 + tan²A and expands to the RHS (1)
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tan 60° = √3, cos 45° = 1/√2, sec 30° = 2/√3, so the numerator is 2(3) + 6(1/2) − 3(4/3) = 6 + 3 − 4 = 5. sin²60° + cos²60° = 3/4 + 1/4 = 1 and cot 30° = √3, so the denominator is 1 + 3 = 4. Value = 5/4.
- numerator = 5 (1)
- denominator = 4 (1)
- value = 5/4 (1)
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