Introduction to Trigonometry Test Class 10 Maths — 40 Marks
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PERIODIC TEST — Introduction to Trigonometry (2026-27) | CLASS 10 | MATHEMATICS
Max Marks: 40 · Time: 1 hour 30 minutes (Time left: 90:00)
Name: ______________________________ Roll No: ____________ Date: ____________
| Section | Marks | Questions | Time | Strategy |
|---|---|---|---|---|
| A | 12 | 12 MCQ | 15 min | Recall the standard-angle values |
| B | 8 | 4 Short | 20 min | Write the ratio, then substitute |
| C | 12 | 4 Long | 30 min | Take the side that is common and simplify |
| D | 8 | 2 Case | 20 min | Draw the right triangle and label the sides |
| Review | — | — | 5 min | Check every identity step |
General Instructions: All questions are compulsory. Section A has 12 MCQs of 1 mark each. Internal choice is given in Q16 of Section B, Q19 and Q20 of Section C, and Q22 of Section D — attempt any ONE of the two options. Give reasons for each step in proofs. Use the exact values of the trigonometric ratios of 0°, 30°, 45°, 60° and 90°. A rough sketch is enough for the case-study parts.
What is this paper? A printable 40-mark Class 10 Maths chapter test on Introduction to Trigonometry with a full per-question marking scheme, internal choice and practice/test modes. It is built to the current CBSE 2026-27 syllabus and is board-pattern practice, not a prediction of any examination paper. Section A is scored for you; use the answer key and the self-assessment box to reach a total out of 40.
sin2θ + cos2θ = 1
1 + tan2θ = sec2θ
1 + cot2θ = cosec2θ
sin(90° − θ) = cos θ
The other two complementary relations follow the same pattern: cos(90° − θ) = sin θ and tan(90° − θ) = cot θ.
Section A — Multiple Choice (12 × 1 = 12 marks)
Choose the single best option. Every solution names the common trap behind the tempting wrong option.
In a right triangle ABC, right-angled at B, if ∠A = θ, then sin θ equals
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Definition of sine
sin θ = (side opposite θ) / hypotenuse
The side opposite ∠A is BC, and the hypotenuse is AC.
sin A = BC / AC
Note: BC / AB is tan A, not sin A.
The side opposite ∠A is BC, and the hypotenuse is AC.
sin A = BC / AC
Note: BC / AB is tan A, not sin A.
Answer: (b) BC / AC
Common trap: Option (c) BC / AB is the ratio of the opposite side to the adjacent side — that is tan A, not sin A.
If sin A = 3/5 and A is an acute angle, then cos A equals
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
First identity
sin2A + cos2A = 1
cos2A = 1 − sin2A
cos2A = 1 − (3/5)2
cos2A = 1 − 9/25
cos2A = 16/25
cos A = 4/5 (A is acute, so cos A > 0)
cos2A = 1 − sin2A
cos2A = 1 − (3/5)2
cos2A = 1 − 9/25
cos2A = 16/25
cos A = 4/5 (A is acute, so cos A > 0)
Answer: (c) 4 / 5
Common trap: Option (a) 5/3 flips the fraction — that is cosec A, not cos A. The positive root 4/5 is correct because A is acute.
The value of sin 30° + cos 60° is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Substitute the standard values
sin 30° = 1/2
cos 60° = 1/2
So sin 30° + cos 60°
= 1/2 + 1/2
= 1
cos 60° = 1/2
So sin 30° + cos 60°
= 1/2 + 1/2
= 1
Answer: (b) 1
Common trap: Option (a) 1/2 stops after one term; option (c) √3/2 is sin 60° (or cos 30°), which is not asked here.
The value of tan 45° + cot 45° is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Substitute the standard values
tan 45° = 1
cot 45° = 1
So tan 45° + cot 45°
= 1 + 1
= 2
cot 45° = 1
So tan 45° + cot 45°
= 1 + 1
= 2
Answer: (d) 2
Common trap: Option (a) 1 is just one of the two values. Option (c) √3 is tan 60°, not tan 45°.
The value of sin 60° × cos 30° is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Substitute and multiply
sin 60° = √3/2
cos 30° = √3/2
So sin 60° × cos 30°
= (√3/2) × (√3/2)
= (3)/(4)
= 3/4
cos 30° = √3/2
So sin 60° × cos 30°
= (√3/2) × (√3/2)
= (3)/(4)
= 3/4
Answer: (a) 3 / 4
Common trap: Option (b) √3/2 is sin 60° alone; multiplying by cos 30° squares the value to 3/4.
If sin A = cos B, where A and B are acute angles, then A + B equals
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Complementary-angle relation
sin A = cos(90° − A)
Given sin A = cos B
So cos(90° − A) = cos B
Since both angles are acute, 90° − A = B
Therefore A + B = 90°
Given sin A = cos B
So cos(90° − A) = cos B
Since both angles are acute, 90° − A = B
Therefore A + B = 90°
Answer: (c) 90°
Common trap: Option (b) 45° is the value of A (or B) when A = B, not the sum. acute angles force A + B = 90°, not 180°.
The value of sin2θ + cos2θ is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
First identity
For every angle θ, sin2θ + cos2θ = 1.
This is a standard identity, true for all values of θ.
This is a standard identity, true for all values of θ.
Answer: (d) 1
Common trap: Option (c) sin 2θ is a different quantity (= 2 sin θ cos θ). The sum of squares is always exactly 1.
The value of sec2θ − tan2θ is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Second identity
1 + tan2θ = sec2θ
Rearranging: sec2θ − tan2θ = 1
So the value is 1 for every angle θ.
Rearranging: sec2θ − tan2θ = 1
So the value is 1 for every angle θ.
Answer: (c) 1
Common trap: Option (a) 0 confuses this with sec2θ − cosec2θ, which is not a constant. sec2θ − tan2θ is fixed at 1.
If tan A = 3/4, then sin A equals
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Build the 3, 4, 5 triangle
tan A = (side opposite A) / (side adjacent to A) = 3/4
Take opposite = 3 and adjacent = 4.
hypotenuse = √(32 + 42) = √(9 + 16) = √25 = 5
sin A = opposite / hypotenuse = 3/5
Take opposite = 3 and adjacent = 4.
hypotenuse = √(32 + 42) = √(9 + 16) = √25 = 5
sin A = opposite / hypotenuse = 3/5
Answer: (a) 3 / 5
Common trap: Option (b) 4/5 is cos A, and option (c) 5/3 is cosec A. sin A always uses the opposite side over the hypotenuse.
The value of cosec 30° is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Reciprocal of sine
cosec 30° = 1 / sin 30°
sin 30° = 1/2
cosec 30° = 1 ÷ (1/2)
= 2
sin 30° = 1/2
cosec 30° = 1 ÷ (1/2)
= 2
Answer: (d) 2
Common trap: Option (a) 1/2 is sin 30°, not cosec 30°. cosec is the reciprocal, so it is 2 (and is always at least 1).
The value of sin 0° + cos 0° is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Substitute the standard values
sin 0° = 0
cos 0° = 1
So sin 0° + cos 0°
= 0 + 1
= 1
cos 0° = 1
So sin 0° + cos 0°
= 0 + 1
= 1
Answer: (b) 1
Common trap: Option (a) 0 remembers only sin 0° and forgets that cos 0° = 1. At θ = 0°, cos θ is at its largest.
If cosec θ = 2, then sin θ equals
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Reciprocal relation
cosec θ = 1 / sin θ
So sin θ = 1 / cosec θ
sin θ = 1 / 2
So sin θ = 1 / cosec θ
sin θ = 1 / 2
Answer: (a) 1 / 2
Common trap: Option (b) 2 repeats cosec θ; sin θ is its reciprocal. Option (c) √3/2 is cos 30°, not related here.
Section B — Short Answer (4 × 2 = 8 marks)
If tan A = 4/3, find sin A and cos A.
Marking Scheme: Correct ratio set-up from a right triangle 1 mark · Both answers correct 1 mark.
Total: 2 Marks
Total: 2 Marks
Set up the right triangle
tan A = (side opposite A) / (side adjacent to A) = 4/3
Take opposite = 4k and adjacent = 3k.
Take opposite = 4k and adjacent = 3k.
Find the hypotenuse
hypotenuse = √((4k)2 + (3k)2)
= √(16k2 + 9k2)
= √(25k2)
= 5k
= √(16k2 + 9k2)
= √(25k2)
= 5k
Write sin A and cos A
sin A = opposite / hypotenuse = 4k / 5k = 4/5
cos A = adjacent / hypotenuse = 3k / 5k = 3/5
cos A = adjacent / hypotenuse = 3k / 5k = 3/5
Answer: sin A = 4/5, cos A = 3/5
Evaluate: sin 30° cos 60° + cos 30° sin 60°
Marking Scheme: All four standard values substituted 1 mark · Correct final value 1 mark.
Total: 2 Marks
Total: 2 Marks
Substitute the standard values
sin 30° = 1/2 · cos 60° = 1/2
cos 30° = √3/2 · sin 60° = √3/2
cos 30° = √3/2 · sin 60° = √3/2
Multiply and add
= (1/2)(1/2) + (√3/2)(√3/2)
= 1/4 + 3/4
= 4/4
= 1
= 1/4 + 3/4
= 4/4
= 1
Answer: 1
If sin θ = cos θ, find θ where 0° < θ < 90°, and hence find the value of 2 sin2θ − 1.
Marking Scheme: Finding θ = 45° 1 mark · Correct value of the expression 1 mark.
Total: 2 Marks
Total: 2 Marks
Find θ
sin θ = cos θ
Divide both sides by cos θ:
tan θ = 1
So θ = 45°
Divide both sides by cos θ:
tan θ = 1
So θ = 45°
Evaluate the expression
2 sin245° − 1
sin 45° = 1/√2, so sin245° = 1/2
= 2(1/2) − 1
= 1 − 1
= 0
sin 45° = 1/√2, so sin245° = 1/2
= 2(1/2) − 1
= 1 − 1
= 0
Answer: θ = 45°; value = 0
Answer either Option 1 or Option 2.
Option 1 — Find cos A and tan A
If sin A = 5/13 and A is an acute angle, find cos A and tan A.
— OR —
Option 2 — Evaluate
Evaluate: 2 sin230° − 3 cos245° + tan260°
Solution — Option 1
Marking Scheme: Correct cos A 1 mark · Correct tan A 1 mark.
Total: 2 Marks
Total: 2 Marks
Use the first identity
cos2A = 1 − sin2A
= 1 − (5/13)2
= 1 − 25/169
= 144/169
cos A = 12/13
= 1 − (5/13)2
= 1 − 25/169
= 144/169
cos A = 12/13
Find tan A
tan A = sin A / cos A
= (5/13) ÷ (12/13)
= 5/12
= (5/13) ÷ (12/13)
= 5/12
Answer: cos A = 12/13, tan A = 5/12
— OR —
Solution — Option 2
Marking Scheme: Standard values substituted 1 mark · Correct final value 1 mark.
Total: 2 Marks
Total: 2 Marks
Substitute the standard values
sin 30° = 1/2, so sin230° = 1/4
cos 45° = 1/√2, so cos245° = 1/2
tan 60° = √3, so tan260° = 3
cos 45° = 1/√2, so cos245° = 1/2
tan 60° = √3, so tan260° = 3
Simplify
= 2(1/4) − 3(1/2) + 3
= 1/2 − 3/2 + 3
= −1 + 3
= 2
= 1/2 − 3/2 + 3
= −1 + 3
= 2
Answer: 2
[Attempt any ONE of the two options of Q16]
Section C — Long Answer (4 × 3 = 12 marks)
Prove that (sin θ)/(1 + cos θ) + (1 + cos θ)/(sin θ) = 2 cosec θ.
Marking Scheme: Taking the LCM and forming one fraction 1 mark · Simplifying the numerator using the identity 1 mark · Final conclusion 1 mark.
Total: 3 Marks
Total: 3 Marks
Take the LCM
LHS = sin θ/(1 + cos θ) + (1 + cos θ)/sin θ
= [sin2θ + (1 + cos θ)2] / [sin θ (1 + cos θ)]
= [sin2θ + (1 + cos θ)2] / [sin θ (1 + cos θ)]
Simplify the numerator
sin2θ + (1 + cos θ)2
= sin2θ + 1 + 2cos θ + cos2θ
= (sin2θ + cos2θ) + 1 + 2cos θ
= 1 + 1 + 2cos θ
= 2 + 2cos θ
= 2(1 + cos θ)
= sin2θ + 1 + 2cos θ + cos2θ
= (sin2θ + cos2θ) + 1 + 2cos θ
= 1 + 1 + 2cos θ
= 2 + 2cos θ
= 2(1 + cos θ)
Cancel the common factor
LHS = 2(1 + cos θ) / [sin θ (1 + cos θ)]
= 2 / sin θ
= 2 cosec θ
= RHS
= 2 / sin θ
= 2 cosec θ
= RHS
Answer: LHS = 2 cosec θ = RHS (hence proved)
If 3 cot A = 4, find the value of (1 − tan2A)/(1 + tan2A).
Marking Scheme: Finding tan A from cot A 1 mark · Substituting the value 1 mark · Correct final value 1 mark.
Total: 3 Marks
Total: 3 Marks
Find tan A
3 cot A = 4
cot A = 4/3
tan A = 1 / cot A = 3/4
cot A = 4/3
tan A = 1 / cot A = 3/4
Substitute into the expression
tan A = 3/4, so tan2A = 9/16
(1 − 9/16) / (1 + 9/16)
= (7/16) / (25/16)
(1 − 9/16) / (1 + 9/16)
= (7/16) / (25/16)
Simplify
= 7/16 × 16/25
= 7/25
= 7/25
Answer: 7/25
Answer either Option 1 or Option 2.
Option 1 — Proof
Prove that (cosec θ − cot θ)2 = (1 − cos θ)/(1 + cos θ).
— OR —
Option 2 — Proof
Prove that sin4θ + cos4θ = 1 − 2 sin2θ cos2θ.
Solution — Option 1
Marking Scheme: Writing cosec and cot in terms of sin and cos 1 mark · Squaring the fraction 1 mark · Using the identity and concluding 1 mark.
Total: 3 Marks
Total: 3 Marks
Convert to sin and cos
cosec θ = 1/sin θ, cot θ = cos θ/sin θ
cosec θ − cot θ = (1 − cos θ)/sin θ
cosec θ − cot θ = (1 − cos θ)/sin θ
Square it
(cosec θ − cot θ)2 = (1 − cos θ)2 / sin2θ
Apply the identity
sin2θ = 1 − cos2θ = (1 − cos θ)(1 + cos θ)
So the expression = (1 − cos θ)2 / [(1 − cos θ)(1 + cos θ)]
= (1 − cos θ)/(1 + cos θ)
So the expression = (1 − cos θ)2 / [(1 − cos θ)(1 + cos θ)]
= (1 − cos θ)/(1 + cos θ)
Answer: (cosec θ − cot θ)2 = (1 − cos θ)/(1 + cos θ) (hence proved)
— OR —
Solution — Option 2
Marking Scheme: Writing the sum as a square 1 mark · Expanding the square 1 mark · Using the identity and concluding 1 mark.
Total: 3 Marks
Total: 3 Marks
Write as a perfect square
sin4θ + cos4θ = (sin2θ)2 + (cos2θ)2
= (sin2θ + cos2θ)2 − 2 sin2θ cos2θ
= (sin2θ + cos2θ)2 − 2 sin2θ cos2θ
Apply the first identity
sin2θ + cos2θ = 1
= (1)2 − 2 sin2θ cos2θ
= (1)2 − 2 sin2θ cos2θ
Conclusion
= 1 − 2 sin2θ cos2θ
= RHS
= RHS
Answer: sin4θ + cos4θ = 1 − 2 sin2θ cos2θ (hence proved)
[Attempt any ONE of the two options of Q19]
Answer either Option 1 or Option 2.
Option 1 — Find sin θ and cos θ
If tan θ + cot θ = 2, find sin θ and cos θ, where 0° < θ < 90°.
— OR —
Option 2 — Proof
Prove that sec2θ + cosec2θ = sec2θ · cosec2θ.
Solution — Option 1
Marking Scheme: Forming the equation in tan θ 1 mark · Solving for tan θ 1 mark · Correct values of sin θ and cos θ 1 mark.
Total: 3 Marks
Total: 3 Marks
Form the equation
cot θ = 1 / tan θ
Let t = tan θ.
t + 1/t = 2
Multiply by t: t2 + 1 = 2t
Let t = tan θ.
t + 1/t = 2
Multiply by t: t2 + 1 = 2t
Solve for t
t2 − 2t + 1 = 0
(t − 1)2 = 0
t = 1
So tan θ = 1 and θ = 45°
(t − 1)2 = 0
t = 1
So tan θ = 1 and θ = 45°
Write sin θ and cos θ
sin 45° = 1/√2
cos 45° = 1/√2
cos 45° = 1/√2
Answer: θ = 45°, sin θ = 1/√2, cos θ = 1/√2
— OR —
Solution — Option 2
Marking Scheme: Writing sec and cosec in terms of sin and cos 1 mark · Taking the common denominator 1 mark · Using the identity and concluding 1 mark.
Total: 3 Marks
Total: 3 Marks
Convert to sin and cos
sec2θ = 1/cos2θ
cosec2θ = 1/sin2θ
cosec2θ = 1/sin2θ
Add with a common denominator
LHS = 1/cos2θ + 1/sin2θ
= (sin2θ + cos2θ) / (sin2θ cos2θ)
= (sin2θ + cos2θ) / (sin2θ cos2θ)
Apply the first identity
sin2θ + cos2θ = 1
LHS = 1 / (sin2θ cos2θ)
= (1/cos2θ) · (1/sin2θ)
= sec2θ cosec2θ
= RHS
LHS = 1 / (sin2θ cos2θ)
= (1/cos2θ) · (1/sin2θ)
= sec2θ cosec2θ
= RHS
Answer: sec2θ + cosec2θ = sec2θ cosec2θ (hence proved)
[Attempt any ONE of the two options of Q20]
Section D — Case Study (2 × 4 = 8 marks)
A right triangle ABC is right-angled at B. Its two legs are AB = 7 cm and BC = 24 cm. Use this triangle to answer the questions below.
(a) Find the length of the hypotenuse AC.
(b) Write the values of sin A and cos A.
(c) Find the value of sin2A + cos2A and name the identity you used.
(a) Find the length of the hypotenuse AC.
(b) Write the values of sin A and cos A.
(c) Find the value of sin2A + cos2A and name the identity you used.
Schematic — not to scale
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Find AC
AC2 = AB2 + BC2
= 72 + 242
= 49 + 576
= 625
AC = √625 = 25 cm
= 72 + 242
= 49 + 576
= 625
AC = √625 = 25 cm
(b) Write sin A and cos A
Side opposite ∠A = BC = 24 cm
Side adjacent to ∠A = AB = 7 cm
Hypotenuse = AC = 25 cm
sin A = 24/25
cos A = 7/25
Side adjacent to ∠A = AB = 7 cm
Hypotenuse = AC = 25 cm
sin A = 24/25
cos A = 7/25
(c) Find sin2A + cos2A
= (24/25)2 + (7/25)2
= 576/625 + 49/625
= 625/625
= 1
Identity used: sin2θ + cos2θ = 1
= 576/625 + 49/625
= 625/625
= 1
Identity used: sin2θ + cos2θ = 1
Answer: (a) AC = 25 cm (b) sin A = 24/25, cos A = 7/25 (c) 1, by the identity sin2θ + cos2θ = 1
Answer either Option 1 or Option 2.
Option 1 — Triangle PQR
In a right triangle PQR, right-angled at Q, PR = 13 cm and PQ = 5 cm.
(a) Find QR.
(b) Write sin P, cos P and tan P.
(c) Find the value of 2 sin P cos P.
(a) Find QR.
(b) Write sin P, cos P and tan P.
(c) Find the value of 2 sin P cos P.
Schematic — not to scale
— OR —
Option 2 — Triangle ABC
In a right triangle ABC, right-angled at B, AB = 8 cm and BC = 15 cm.
(a) Find AC.
(b) Write sin A, cos A and tan A.
(c) Find the value of (sin A)/(cos A) and state what it equals.
(a) Find AC.
(b) Write sin A, cos A and tan A.
(c) Find the value of (sin A)/(cos A) and state what it equals.
Schematic — not to scale
Solution — Option 1
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Find QR
PR2 = PQ2 + QR2
QR2 = 132 − 52
= 169 − 25
= 144
QR = 12 cm
QR2 = 132 − 52
= 169 − 25
= 144
QR = 12 cm
(b) Write the ratios
Opposite ∠P = QR = 12 cm
Adjacent to ∠P = PQ = 5 cm
Hypotenuse = PR = 13 cm
sin P = 12/13
cos P = 5/13
tan P = 12/5
Adjacent to ∠P = PQ = 5 cm
Hypotenuse = PR = 13 cm
sin P = 12/13
cos P = 5/13
tan P = 12/5
(c) Find 2 sin P cos P
= 2 × (12/13) × (5/13)
= (2 × 12 × 5) / (13 × 13)
= 120/169
= (2 × 12 × 5) / (13 × 13)
= 120/169
Answer: (a) QR = 12 cm (b) sin P = 12/13, cos P = 5/13, tan P = 12/5 (c) 120/169
— OR —
Solution — Option 2
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Find AC
AC2 = AB2 + BC2
= 82 + 152
= 64 + 225
= 289
AC = √289 = 17 cm
= 82 + 152
= 64 + 225
= 289
AC = √289 = 17 cm
(b) Write the ratios
Opposite ∠A = BC = 15 cm
Adjacent to ∠A = AB = 8 cm
Hypotenuse = AC = 17 cm
sin A = 15/17
cos A = 8/17
tan A = 15/8
Adjacent to ∠A = AB = 8 cm
Hypotenuse = AC = 17 cm
sin A = 15/17
cos A = 8/17
tan A = 15/8
(c) Find sin A / cos A
= (15/17) ÷ (8/17)
= (15/17) × (17/8)
= 15/8
This equals tan A.
= (15/17) × (17/8)
= 15/8
This equals tan A.
Answer: (a) AC = 17 cm (b) sin A = 15/17, cos A = 8/17, tan A = 15/8 (c) sin A / cos A = 15/8 = tan A
[Attempt any ONE of the two options of Q22]
Answer Key & Marks (tap to open)
| Q | Section | Marks | Answer |
|---|---|---|---|
| 1 | A | 1 | (b) BC / AC |
| 2 | A | 1 | (c) 4/5 |
| 3 | A | 1 | (b) 1 |
| 4 | A | 1 | (d) 2 |
| 5 | A | 1 | (a) 3/4 |
| 6 | A | 1 | (c) 90° |
| 7 | A | 1 | (d) 1 |
| 8 | A | 1 | (c) 1 |
| 9 | A | 1 | (a) 3/5 |
| 10 | A | 1 | (d) 2 |
| 11 | A | 1 | (b) 1 |
| 12 | A | 1 | (a) 1/2 |
| 13–16 | B | 8 | Written — see the marking scheme under each question (Q16 has a choice) |
| 17–20 | C | 12 | Written — see the marking scheme under each question (Q19, Q20 have a choice) |
| 21–22 | D | 8 | Written — see the marking scheme under each question (Q22 has a choice) |
| Total | 40 | Section A 12 + Sections B–D 28 | |
Self-assessment for Sections B, C and D (28 marks)
The MCQ section is scored for you. For the written questions, tick each one you got right out of the marks shown, and add them up honestly.
Written total: 0 / 28
Paper total: score Section A above, then add your 0 written marks for the total out of 40.
Questions students ask about this paper
- Is this the actual CBSE board question paper?
- No. It is original board-pattern practice built to the CBSE Class 10 Maths syllabus 2026-27 (NCERT Chapter 8, Introduction to Trigonometry). It is not a prediction of any examination paper.
- How is the paper scored?
- Section A (12 MCQs) is scored automatically when you press Submit & Score. Sections B, C and D are written, so you check them against the marking scheme under each question and tick the self-assessment box; the box then shows your total out of 40.
- What does the test cover?
- Trigonometric ratios, the values of the ratios at 0°, 30°, 45°, 60° and 90°, ratios of complementary angles, and the three basic identities — exactly the scope of NCERT Chapter 8.
- Do my answers stay if I reload the page?
- Yes. Your selected options and your mode are kept for this browser tab, so an interrupted session resumes where you left off.
Keep going with these
Written and checked by Pandey Sir · Pandey Sir’s BEST Classes, Dammaiguda, Hyderabad. Every answer on this paper is worked out and checked before publishing.
This is board-pattern practice for Introduction to Trigonometry, built to the current CBSE Class 10 Maths syllabus 2026-27. It is not a prediction of any examination paper.
Curriculum scope checked against cbseacademic.nic.in. Spotted a mistake? Tell us here.