CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 24

Download CBSE MCQs for Class 10 Mathematics: Chapter 08 Introduction to Trigonometry

Review structured MCQ sets for Class 10 Mathematics Chapter 08 Introduction to Trigonometry. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Chapter 08 Introduction to Trigonometry

Navigate directly to the 50 objective questions for Chapter 08 Introduction to Trigonometry using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: In a right triangle ABC, right-angled at B, AB = 3 cm and AC = 5 cm. For the angle C, sin C, cos C, tan C are:
(a) \( \frac35,\ \frac45,\ \frac34 \)
(b) \( \frac45,\ \frac35,\ \frac43 \)
(c) \( \frac35,\ \frac45,\ \frac43 \)
(d) \( \frac34,\ \frac45,\ \frac35 \)
Show Answer & Explanation

Answer: (a) \( \frac35,\ \frac45,\ \frac34 \)

Explanation:
1. BC = sqrt(25 - 9) = 4 cm. For angle C: opposite side AB = 3, adjacent BC = 4, hypotenuse AC = 5. So sin C = 3/5, cos C = 4/5, tan C = 3/4.

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 2: If tan theta = m/n, then sin theta equals:
(a) \( \frac{n}{\sqrt{m^2+n^2}} \)
(b) \( \frac{m}{m+n} \)
(c) \( \frac{m}{\sqrt{m^2-n^2}} \)
(d) \( \frac{m}{\sqrt{m^2+n^2}} \)
Show Answer & Explanation

Answer: (d) \( \frac{m}{\sqrt{m^2+n^2}} \)

Explanation:
1. Take perpendicular = m and base = n. Hypotenuse = sqrt(m^2 + n^2), so sin theta = m / sqrt(m^2 + n^2).

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 3: If cosec A = 13/5, then tan^2 A - sin^2 A equals:
(a) \( \frac{25}{144} \)
(b) \( \frac{25}{169} \)
(c) \( \frac{625}{24336} \)
(d) \( \frac{625}{169} \)
Show Answer & Explanation

Answer: (c) \( \frac{625}{24336} \)

Explanation:
1. With perpendicular 5 and hypotenuse 13, the base is 12. tan^2 A - sin^2 A = 25/144 - 25/169 = 25(169 - 144)/(144 x 169) = 625/24336. This equals sin^4 A sec^2 A = (625/28561)(169/144).

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 4: In triangle ABC right-angled at B, AC + AB = 9 cm and BC = 3 cm. Then cot C, cosec C, sec C are:
(a) \( \frac43,\ \frac54,\ \frac53 \)
(b) \( \frac34,\ \frac54,\ \frac53 \)
(c) \( \frac34,\ \frac53,\ \frac54 \)
(d) \( \frac34,\ \frac45,\ \frac35 \)
Show Answer & Explanation

Answer: (b) \( \frac34,\ \frac54,\ \frac53 \)

Explanation:
1. Put AC = 9 - AB: (9 - AB)^2 = AB^2 + 9 gives 72 = 18 AB, so AB = 4 and AC = 5. For angle C: cot C = BC/AB = 3/4, cosec C = AC/AB = 5/4, sec C = AC/BC = 5/3.

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 5: The value of cos 15 degrees is:
(a) \( \frac{\sqrt3-1}{2\sqrt2} \)
(b) \( \frac{\sqrt3+1}{2\sqrt2} \)
(c) \( \frac{\sqrt3+1}{2} \)
(d) \( \frac{1}{2\sqrt2} \)
Show Answer & Explanation

Answer: (b) \( \frac{\sqrt3+1}{2\sqrt2} \)

Explanation:
1. cos(45 - 30) = cos 45 cos 30 + sin 45 sin 30 = (1/sqrt2)(sqrt3/2) + (1/sqrt2)(1/2) = (sqrt3 + 1)/(2 sqrt2).

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 6: A rhombus of side 10 cm has two angles of 60 degrees. The diagonals and the area are:
(a) \( 10\ \text{cm and }10\sqrt3\ \text{cm; area }50\sqrt3\ \text{cm}^2 \)
(b) \( 10\ \text{cm and }5\sqrt3\ \text{cm; area }25\sqrt3\ \text{cm}^2 \)
(c) \( 20\ \text{cm and }10\sqrt3\ \text{cm; area }100\sqrt3\ \text{cm}^2 \)
(d) \( 10\sqrt3\ \text{cm each; area }150\ \text{cm}^2 \)
Show Answer & Explanation

Answer: (a) \( 10\ \text{cm and }10\sqrt3\ \text{cm; area }50\sqrt3\ \text{cm}^2 \)

Explanation:
1. The diagonal BD bisects the 60 degree angles, so triangle ABD is equilateral and BD = 10. In the right triangle AOB, OA = 10 cos 30 = 5 sqrt3, so AC = 10 sqrt3. Area = (1/2)(10 sqrt3)(10) = 50 sqrt3 cm^2.

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 7: The value of (sec^2 54 - cot^2 36)/(cosec^2 57 - tan^2 33) + 2 sin^2 38 sec^2 52 - sin^2 45 + (2/sqrt3) tan 17 tan 60 tan 73 is:
(a) \( \frac52 \)
(b) \( \frac{11}{2} \)
(c) \( 4 \)
(d) \( \frac92 \)
Show Answer & Explanation

Answer: (d) \( \frac92 \)

Explanation:
1. Since sec 54 = cosec 36 and cosec 57 = sec 33, the first fraction is (cosec^2 36 - cot^2 36)/(sec^2 33 - tan^2 33) = 1. sin 38 sec 52 = 1 gives 2. sin^2 45 = 1/2. tan 17 tan 73 = 1 gives (2/sqrt3)(sqrt3) = 2. Total 1 + 2 - 1/2 + 2 = 9/2.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 8: The value of cosec(65 + theta) - sec(25 - theta) - tan(55 - theta) + cot(35 + theta) is:
(a) 1
(b) 2
(c) 0
(d) -1
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. cosec(65 + theta) = sec(25 - theta) and cot(35 + theta) = tan(55 - theta) because the angles are complementary. The expression becomes zero.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 9: cot theta - tan theta equals:
(a) \( \frac{2\sin^2\theta-1}{\sin\theta\cos\theta} \)
(b) \( \frac{1}{\sin\theta\cos\theta} \)
(c) \( \frac{2\cos^2\theta-1}{\sin\theta\cos\theta} \)
(d) \( \frac{\cos 2\theta+1}{\sin\theta} \)
Show Answer & Explanation

Answer: (c) \( \frac{2\cos^2\theta-1}{\sin\theta\cos\theta} \)

Explanation:
1. cot theta - tan theta = (cos^2 theta - sin^2 theta)/(sin theta cos theta) = (2cos^2 theta - 1)/(sin theta cos theta), using sin^2 = 1 - cos^2.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 10: The value of (cosec A - sin A)(sec A - cos A)(tan A + cot A) is:
(a) 0
(b) 1
(c) sin A cos A
(d) 2
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. cosec A - sin A = cos^2 A / sin A and sec A - cos A = sin^2 A / cos A, while tan A + cot A = 1/(sin A cos A). The product is (cos^2 A sin^2 A)/(sin A cos A) x 1/(sin A cos A) = 1.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 11: If sin theta + cos theta = m and sec theta + cosec theta = n, then n(m^2 - 1) equals:
(a) 2m
(b) m
(c) 2n
(d) m^2
Show Answer & Explanation

Answer: (a) 2m

Explanation:
1. m^2 - 1 = 2 sin theta cos theta and n = (sin theta + cos theta)/(sin theta cos theta) = m/(sin theta cos theta). So n(m^2 - 1) = 2m.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 12: If sec theta = x + 1/(4x) with x = 1, then sec theta + tan theta equals:
(a) \( 2\ \text{only} \)
(b) \( \frac12\ \text{only} \)
(c) \( \frac54\ \text{or}\ \frac34 \)
(d) \( 2\ \text{or}\ \frac12 \)
Show Answer & Explanation

Answer: (d) \( 2\ \text{or}\ \frac12 \)

Explanation:
1. sec theta = 5/4, so tan theta = +/- 3/4. Then sec + tan = 2 or 1/2, which equal 2x and 1/(2x).

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 13: If theta is acute and tan theta + cot theta = 2, then tan^9 theta + cot^9 theta equals:
(a) 1
(b) 0
(c) 512
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. tan theta + 1/tan theta = 2 gives (tan theta - 1)^2 = 0, so tan theta = 1 and theta = 45 degrees. Then tan^9 + cot^9 = 1 + 1 = 2.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 14: If sin alpha = 1/3 and alpha + beta = 90 degrees, then sin beta equals:
(a) \( \frac{\sqrt2}{3} \)
(b) \( \frac13 \)
(c) \( \frac{2\sqrt2}{3} \)
(d) \( \frac34 \)
Show Answer & Explanation

Answer: (c) \( \frac{2\sqrt2}{3} \)

Explanation:
1. sin beta = sin(90 - alpha) = cos alpha = sqrt(1 - 1/9) = 2 sqrt2 / 3.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 15: If 5 tan theta = 4, then (5 sin theta - 3 cos theta)/(5 sin theta + 2 cos theta) equals:
(a) \( \frac13 \)
(b) \( \frac16 \)
(c) \( \frac45 \)
(d) \( \frac23 \)
Show Answer & Explanation

Answer: (b) \( \frac16 \)

Explanation:
1. Divide numerator and denominator by cos theta: (5 tan theta - 3)/(5 tan theta + 2) = (4 - 3)/(4 + 2) = 1/6.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 16: If 7 sin alpha = 24 cos alpha with alpha acute, then 14 tan alpha - 75 cos alpha - 7 sec alpha equals:
(a) 2
(b) 1
(c) 3
(d) 4
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. tan alpha = 24/7 so the triangle is 7, 24, 25: cos alpha = 7/25 and sec alpha = 25/7. 14(24/7) - 75(7/25) - 7(25/7) = 48 - 21 - 25 = 2.

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 17: If 3 sin beta + 5 cos beta = 5, then (3 cos beta - 5 sin beta)^2 equals:
(a) 9
(b) 1
(c) 25
(d) 34
Show Answer & Explanation

Answer: (a) 9

Explanation:
1. (3 sin + 5 cos)^2 + (3 cos - 5 sin)^2 = 9 + 25 = 34. So (3 cos beta - 5 sin beta)^2 = 34 - 25 = 9.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 18: If tan theta = 4, then tan theta / (sin^3 theta / cos theta + sin theta cos theta) equals:
(a) 0
(b) sqrt2
(c) 4
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. sin^3/cos + sin cos = sin(sin^2 + cos^2)/cos = tan theta. So the ratio is 1 (the value of tan theta does not matter).

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 19: The value of tan 5 tan 10 tan 15 tan 20 ... tan 85 (angles in degrees) is:
(a) 2
(b) 3
(c) 1
(d) 0
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. Pair tan x with tan(90 - x) = cot x, whose product is 1. The middle term is tan 45 = 1, so the whole product is 1.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 20: As x increases from 0 to pi/2, the value of cos x:
(a) increases
(b) decreases
(c) remains constant
(d) increases and then decreases
Show Answer & Explanation

Answer: (b) decreases

Explanation:
1. cos 0 = 1 and cos(pi/2) = 0, and cos x falls steadily between them.

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 21: The value of x satisfying x sin(pi/6) cos^2(pi/4) = cot^2(pi/6) sec(pi/3) tan(pi/4) / (cosec^2(pi/4) cosec(pi/6)) is:
(a) 4
(b) 6
(c) -2
(d) 0
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. Left side: x (1/2)(1/2) = x/4. Right side: (3)(2)(1)/((2)(2)) = 3/2. So x/4 = 3/2 and x = 6.

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 22: The area of a triangle is 12 square cm and two sides are 6 cm and 12 cm. The included angle is:
(a) \( \sin^{-1}\left(\frac13\right) \)
(b) \( \sin^{-1}\left(\frac16\right) \)
(c) \( \cos^{-1}\left(\frac13\right) \)
(d) \( \cos^{-1}\left(\frac16\right) \)
Show Answer & Explanation

Answer: (a) \( \sin^{-1}\left(\frac13\right) \)

Explanation:
1. Area = (1/2)(6)(12) sin theta = 12 gives sin theta = 1/3, so theta = sin^-1(1/3).

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 23: If alpha + beta = 90 degrees and alpha = 2 beta, then cos^2 alpha + sin^2 beta equals:
(a) \( 0 \)
(b) \( 1 \)
(c) \( 2 \)
(d) \( \frac12 \)
Show Answer & Explanation

Answer: (d) \( \frac12 \)

Explanation:
1. 3 beta = 90 so beta = 30 and alpha = 60. cos^2 60 + sin^2 30 = 1/4 + 1/4 = 1/2.

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

Question 24: The value of (sin^2 27 + sin^2 63)/(cos^2 40 + cos^2 50) + sin x cos x sin(90 - x)/cos(90 - x) + cos x sin x cos(90 - x)/sin(90 - x) is:
(a) 1
(b) 0
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. sin^2 27 + sin^2 63 = 1 and cos^2 40 + cos^2 50 = 1, so the first part is 1. The next two terms are sin x cos x cos x / sin x = cos^2 x and cos x sin x sin x / cos x = sin^2 x, which add to 1. Total = 2.

Teacher's Note:
1. Identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2. Complementary angles: sin(90 - A) = cos A, tan(90 - A) = cot A, sec(90 - A) = cosec A. Also tan A cot A = 1.

Question 25: The value of cos 10 cos 20 cos 30 ... cos 180 (angles in degrees) is:
(a) 1
(b) -1
(c) 0
(d) 1/2
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. The product contains cos 90 = 0, so the whole product is 0.

Teacher's Note:
1. In a right triangle sin = perpendicular/hypotenuse, cos = base/hypotenuse, tan = perpendicular/base, and cosec, sec, cot are their reciprocals. Standard angles: sin 30 = 1/2, sin 45 = 1/sqrt2, sin 60 = sqrt3/2, with cos, tan from them.

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Chapter 08 Introduction to Trigonometry Objective Questions & Solutions for Class 10 Mathematics

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