CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 20

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 08 Introduction to Trigonometry

Explore reliable objective questions for Chapter 08 Introduction to Trigonometry tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Chapter 08 Introduction to Trigonometry MCQs for Class 10 Mathematics

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Multiple Choice Questions

Question 1: If \( \operatorname{cosec}\theta = \tfrac{13}{12} \) (θ acute), the value of \( \frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta} \) is:
(a) 1/3
(b) 3
(c) 9
(d) −3
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. sin θ = 12/13 and cos θ = 5/13. The expression is (24 − 15)/(48 − 45) = 9/3 = 3.

Teacher's Note:
1. Find sin and cos from the cosec value.
2. Substitute and divide the common 1/13.

Question 2: The value of \( \frac{\cos 35^\circ}{\sin 55^\circ}+\frac{\tan 27^\circ\tan 63^\circ}{\sin 30^\circ}-3\tan^2 60^\circ \) is:
(a) −6
(b) 6
(c) −3
(d) 0
Show Answer & Explanation

Answer: (a) −6

Explanation:
1. cos 35° = sin 55°, so the first term is 1. tan 27° tan 63° = 1, so the second term is 1/(1/2) = 2. The last term is 3 × 3 = 9. Total 1 + 2 − 9 = −6.

Teacher's Note:
1. Use complementary angles.
2. tan x tan(90° − x) = 1.

Question 3: The value of \( \frac{\sin 39^\circ}{\cos 51^\circ}+2\tan 11^\circ\tan 31^\circ\tan 45^\circ\tan 59^\circ\tan 79^\circ-3(\sin^2 21^\circ+\sin^2 69^\circ) \) is:
(a) 1
(b) 2
(c) −3
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. sin 39° = cos 51°, so the first term is 1. tan 11° tan 79° = 1 and tan 31° tan 59° = 1, tan 45° = 1, so the second term is 2. sin²69° = cos²21°, so the bracket is 1 and the last term is 3. Total 1 + 2 − 3 = 0.

Teacher's Note:
1. Pair angles that add to 90°.
2. sin²21° + cos²21° = 1.

Question 4: \( \frac{1}{\sec x-\tan x}-\frac{1}{\cos x} \) is equal to:
(a) \( \sec x \)
(b) \( \cot x \)
(c) \( \tan x \)
(d) \( \sin x \)
Show Answer & Explanation

Answer: (c) \( \tan x \)

Explanation:
1. 1/(sec x − tan x) = sec x + tan x because (sec x + tan x)(sec x − tan x) = 1. Subtracting sec x leaves tan x.

Teacher's Note:
1. Multiply and divide by sec x + tan x.
2. Recall that 1/cos x = sec x.

Question 5: \( \frac{\sin\theta}{1+\cos\theta}+\frac{1+\cos\theta}{\sin\theta} \) is equal to:
(a) \( 2\sec\theta \)
(b) \( 2\operatorname{cosec}\theta \)
(c) \( 2\sin\theta \)
(d) \( \operatorname{cosec}\theta \)
Show Answer & Explanation

Answer: (b) \( 2\operatorname{cosec}\theta \)

Explanation:
1. Combine: (sin²θ + (1 + cos θ)²)/(sin θ (1 + cos θ)) = (2 + 2cos θ)/(sin θ (1 + cos θ)) = 2/sin θ.

Teacher's Note:
1. Take the common denominator.
2. sin²θ + cos²θ = 1.

Question 6: \( \frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta} \) is equal to:
(a) \( 2\sec\theta \)
(b) \( 2\operatorname{cosec}\theta \)
(c) \( 2\cos\theta \)
(d) \( \sec\theta \)
Show Answer & Explanation

Answer: (a) \( 2\sec\theta \)

Explanation:
1. Combine: ((1 + sin θ)² + cos²θ)/(cos θ (1 + sin θ)) = (2 + 2 sin θ)/(cos θ (1 + sin θ)) = 2/cos θ.

Teacher's Note:
1. Take the common denominator.
2. Use sin²θ + cos²θ = 1.

Question 7: \( \frac{1+\sec\theta}{\sec\theta} \) is equal to:
(a) \( \frac{\cos^2\theta}{1-\cos\theta} \)
(b) \( \frac{\sin^2\theta}{1+\cos\theta} \)
(c) \( \frac{1-\cos\theta}{\sin^2\theta} \)
(d) \( \frac{\sin^2\theta}{1-\cos\theta} \)
Show Answer & Explanation

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

Explanation:
1. (1 + sec θ)/sec θ = cos θ + 1. And sin²θ/(1 − cos θ) = (1 − cos θ)(1 + cos θ)/(1 − cos θ) = 1 + cos θ.

Teacher's Note:
1. Write sec θ = 1/cos θ.
2. Factor sin²θ as (1 − cos θ)(1 + cos θ).

Question 8: \( \frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A} \) is equal to:
(a) \( \sin A-\cos A \)
(b) \( \sin A\cos A \)
(c) \( \sin A+\cos A \)
(d) 1
Show Answer & Explanation

Answer: (c) \( \sin A+\cos A \)

Explanation:
1. cos A/(1 − tan A) = cos²A/(cos A − sin A) and sin A/(1 − cot A) = sin²A/(sin A − cos A) = −sin²A/(cos A − sin A). The sum is (cos²A − sin²A)/(cos A − sin A) = cos A + sin A.

Teacher's Note:
1. Express tan and cot in sin and cos.
2. Use a² − b² = (a − b)(a + b).

Question 9: \( (\cot\theta-\operatorname{cosec}\theta)^2 \) is equal to:
(a) \( \frac{1+\cos\theta}{1-\cos\theta} \)
(b) \( \frac{1-\cos\theta}{1+\cos\theta} \)
(c) \( \frac{1-\sin\theta}{1+\sin\theta} \)
(d) \( \frac{\cos\theta}{1+\cos\theta} \)
Show Answer & Explanation

Answer: (b) \( \frac{1-\cos\theta}{1+\cos\theta} \)

Explanation:
1. (cos θ − 1)²/sin²θ = (1 − cos θ)²/((1 − cos θ)(1 + cos θ)) = (1 − cos θ)/(1 + cos θ).

Teacher's Note:
1. Write over sin θ.
2. Replace sin²θ by (1 − cos θ)(1 + cos θ).

Question 10: The solutions of \( 2\cos^2\theta-\cos\theta=0 \) for \( 0^\circ\le\theta\le90^\circ \) are:
(a) 60° and 90°
(b) 60° only
(c) 30° and 90°
(d) 0° and 60°
Show Answer & Explanation

Answer: (a) 60° and 90°

Explanation:
1. cos θ (2cos θ − 1) = 0 gives cos θ = 0 or cos θ = 1/2, that is θ = 90° or 60°.

Teacher's Note:
1. Factor out cos θ.
2. Both roots lie in the range.

Question 11: The solution of \( 3\sec^2\theta=2\operatorname{cosec}\theta \) for \( 0^\circ<\theta<90^\circ \) is:
(a) 45°
(b) 60°
(c) 15°
(d) 30°
Show Answer & Explanation

Answer: (d) 30°

Explanation:
1. 3/cos²θ = 2/sin θ gives 3 sin θ = 2 cos²θ = 2 − 2 sin²θ, so 2s² + 3s − 2 = 0, (2s − 1)(s + 2) = 0. s = 1/2 and θ = 30°.

Teacher's Note:
1. Cross-multiply and use cos² = 1 − sin².
2. Reject sin θ = −2.

Question 12: The solutions of \( 2\sin^2\theta-3\sin\theta+1=0 \) for \( 0^\circ\le\theta\le90^\circ \) are:
(a) 30° and 60°
(b) 45° and 90°
(c) 30° and 90°
(d) 60° only
Show Answer & Explanation

Answer: (c) 30° and 90°

Explanation:
1. (2 sin θ − 1)(sin θ − 1) = 0 gives sin θ = 1/2 or 1, so θ = 30° or 90°.

Teacher's Note:
1. Factor the quadratic in sin θ.

Question 13: The value of \( \sin 15^\circ \) is:
(a) \( \frac{\sqrt3+1}{2\sqrt2} \)
(b) \( \frac{\sqrt3-1}{2\sqrt2} \)
(c) \( \frac{\sqrt3-1}{2} \)
(d) \( \frac{\sqrt3+1}{2} \)
Show Answer & Explanation

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

Explanation:
1. sin 15° = sin(45° − 30°) = (1/√2)(√3/2) − (1/√2)(1/2) = (√3 − 1)/(2√2).

Teacher's Note:
1. Write 15° = 45° − 30°.
2. Use sin(A − B).

Question 14: If \( \cos(A+B)=\tfrac12 \) and \( \sin(A-B)=\tfrac12 \), with \( 0\lt A+B\le90^\circ \) and A > B, then:
(a) A = 60°, B = 30°
(b) A = 30°, B = 15°
(c) A = 45°, B = 30°
(d) A = 45°, B = 15°
Show Answer & Explanation

Answer: (d) A = 45°, B = 15°

Explanation:
1. A + B = 60° and A − B = 30°, so A = 45° and B = 15°.

Teacher's Note:
1. Convert each equation to an angle.
2. Add and subtract.

Question 15: If \( \sin(A+B)=1 \) and \( \sin(A-B)=\tfrac12 \), with \( 0\lt A+B\le90^\circ \) and A > B, then:
(a) A = 45°, B = 15°
(b) A = 45°, B = 45°
(c) A = 60°, B = 30°
(d) A = 90°, B = 30°
Show Answer & Explanation

Answer: (c) A = 60°, B = 30°

Explanation:
1. A + B = 90° and A − B = 30°, so A = 60° and B = 30°.

Teacher's Note:
1. sin 90° = 1 and sin 30° = 1/2.

Question 16: If \( \tan(A-B)=\tfrac1{\sqrt3} \) and \( \tan(A+B)=\sqrt3 \), with \( 0\lt A+B\le90^\circ \) and A > B, then:
(a) A = 60°, B = 30°
(b) A = 45°, B = 15°
(c) A = 45°, B = 30°
(d) A = 30°, B = 15°
Show Answer & Explanation

Answer: (b) A = 45°, B = 15°

Explanation:
1. A − B = 30° and A + B = 60°, so A = 45° and B = 15°.

Teacher's Note:
1. Recall tan 30° = 1/√3 and tan 60° = √3.

Question 17: The value of \( \frac{\sec^2 54^\circ-\cot^2 36^\circ}{\operatorname{cosec}^2 57^\circ-\tan^2 33^\circ}+2\sin^2 38^\circ\sec^2 52^\circ-\sin^2 45^\circ \) is:
(a) 5/2
(b) 3/2
(c) 7/2
(d) 2
Show Answer & Explanation

Answer: (a) 5/2

Explanation:
1. sec²54° − cot²36° = sec²54° − tan²54° = 1 and cosec²57° − tan²33° = cosec²57° − cot²57° = 1, so the fraction is 1. sec 52° = cosec 38°, so 2 sin²38° sec²52° = 2. Then 1 + 2 − 1/2 = 5/2.

Teacher's Note:
1. Reduce each pair using complementary angles.
2. sec² − tan² = 1 and cosec² − cot² = 1.

Question 18: The angles of a triangle are in A.P. The number of degrees in the least angle is to the number of radians in the greatest angle as 60 : π. The angles are:
(a) 20°, 60°, 100°
(b) 40°, 60°, 80°
(c) 45°, 60°, 75°
(d) 30°, 60°, 90°
Show Answer & Explanation

Answer: (d) 30°, 60°, 90°

Explanation:
1. The middle angle is 60° (sum 180°, in A.P.). Greatest G degrees = Gπ/180 radians, so L/(Gπ/180) = 60/π gives G = 3L. The other two angles add to 120°, so L + 3L = 120, giving L = 30° and G = 90°.

Teacher's Note:
1. In an A.P. of three angles the middle one is the average.
2. Convert the greatest angle to radians.

Question 19: The angles of a triangle are in A.P. The number of grades in the least angle is to the number of radians in the greatest angle as 40 : π. The angles in degrees are:
(a) 30°, 60°, 90°
(b) 40°, 60°, 80°
(c) 20°, 60°, 100°
(d) 10°, 60°, 110°
Show Answer & Explanation

Answer: (c) 20°, 60°, 100°

Explanation:
1. The middle angle is 60°. Least L degrees = 10L/9 grades. Greatest G degrees = Gπ/180 radians. (10L/9)/(Gπ/180) = 40/π gives G = 5L. L + 5L = 120, so L = 20° and G = 100°.

Teacher's Note:
1. The middle angle is 60°.
2. Convert both to the stated units.

Question 20: Arcs of the same length in two circles subtend angles of 60° and 75° at their centres. The ratio of their radii is:
(a) 5 : 4
(b) 4 : 5
(c) 3 : 4
(d) 6 : 5
Show Answer & Explanation

Answer: (a) 5 : 4

Explanation:
1. l = r₁θ₁ = r₂θ₂ gives r₁ : r₂ = θ₂ : θ₁ = 75 : 60 = 5 : 4.

Teacher's Note:
1. The radius is inversely proportional to the angle for a fixed arc.

Question 21: A pendulum of length 50 cm describes an arc of length 10 cm at its tip. The angle through which it swings is (use π = 22/7):
(a) 11° 17′ 26″
(b) 12° 27′ 16″
(c) 10° 27′ 16″
(d) 11° 27′ 16″
Show Answer & Explanation

Answer: (d) 11° 27′ 16″

Explanation:
1. θ = l/r = 10/50 = 0.2 radian = 0.2 × 180 × 7/22 degrees = 11.4545° = 11° 27′ 16″.

Teacher's Note:
1. Use l = rθ.
2. Convert the decimal part into minutes and seconds.

Question 22: A horse tied to a post by a rope moves along a circular path with the rope tight. It describes 88 m when it has turned through 72° at the centre. The length of the rope is (π = 22/7):
(a) 60 m
(b) 77 m
(c) 70 m
(d) 35 m
Show Answer & Explanation

Answer: (c) 70 m

Explanation:
1. 72° = 2π/5 radian. r = 88/(2π/5) = 88 × 5 × 7/(2 × 22) = 70 m.

Teacher's Note:
1. Convert 72° to radians first.

Question 23: The angular diameter of the moon is 30′. A coin of diameter 2.2 cm hides the moon when kept at a distance of (π = 22/7):
(a) 126 cm
(b) 252 cm
(c) 220 cm
(d) 504 cm
Show Answer & Explanation

Answer: (b) 252 cm

Explanation:
1. 30′ = 0.5° = π/360 radian. Distance = 2.2/(π/360) = 2.2 × 360 × 7/22 = 252 cm.

Teacher's Note:
1. Use diameter = distance × angle in radians.

Question 24: A person can read letters that subtend an angle of 5′ at the eye. The height of the letters he can read at a distance of 12 m is about (use π = 3.14):
(a) 1.7 cm
(b) 17 cm
(c) 0.17 cm
(d) 3.4 cm
Show Answer & Explanation

Answer: (a) 1.7 cm

Explanation:
1. 5′ = 5π/(60 × 180) = 0.001454 radian. Height = 1200 cm × 0.001454 = 1.745 cm, about 1.7 cm.

Teacher's Note:
1. Convert 5′ to radians.
2. Height = distance × angle.

Question 25: The angle between the minute hand and the hour hand of a clock at 7:20 is:
(a) 110°
(b) 90°
(c) 120°
(d) 100°
Show Answer & Explanation

Answer: (d) 100°

Explanation:
1. At 7:20 the hour hand is at 7 × 30 + 20 × 0.5 = 220° and the minute hand at 20 × 6 = 120°. The difference is 100°.

Teacher's Note:
1. Hour hand moves 0.5° per minute.
2. Minute hand moves 6° per minute.

Question 26: If \( n\tan\theta=m \), then \( \frac{m\sin\theta+n\cos\theta}{m\sin\theta-n\cos\theta} \) equals:
(a) \( \frac{m^2-n^2}{m^2+n^2} \)
(b) \( \frac{m^2+n^2}{m^2-n^2} \)
(c) \( \frac{m+n}{m-n} \)
(d) \( \frac{m^2+n^2}{mn} \)
Show Answer & Explanation

Answer: (b) \( \frac{m^2+n^2}{m^2-n^2} \)

Explanation:
1. Divide by cos θ: (m tan θ + n)/(m tan θ − n) with tan θ = m/n gives (m²/n + n)/(m²/n − n) = (m² + n²)/(m² − n²).

Teacher's Note:
1. Divide numerator and denominator by cos θ.

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