CBSE Class 10 · Chapter 8 · Trigonometry · 2026-27

Introduction to Trigonometry Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceTrigonometric ratios

In $△ A B C$ , right-angled at $B$ , $A B = 5$ cm and $B C = 12$ cm. Then $\sin A$ is

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Answer: (b)

\frac{12}{13}

Why: sin A = opposite/hypotenuse = BC/AC, with AC = 13.

$A C = \sqrt{25 + 144} = 13$ cm. $\sin A = \frac{B C}{A C} = \frac{12}{13}$ .

Q2

·1 mark·Multiple choiceTrigonometric ratios

If $\tan A = \frac{8}{15}$ , where $A$ is acute, then $\sec A$ equals

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Answer: (a)

\frac{17}{15}

Why: Opposite 8, adjacent 15, hypotenuse 17; sec A = hypotenuse/adjacent.

Take opposite $= 8 k$ , adjacent $= 15 k$ ; hypotenuse $= 17 k$ . $\sec A = \frac{17}{15}$ (or $\sec^{2} A = 1 + \frac{64}{225} = \frac{289}{225}$ ).

Q3

·1 mark·Multiple choiceRatios of standard angles

The value of $\sin 30^{\circ} \cos 60^{\circ} + \cos 30^{\circ} \sin 60^{\circ}$ is

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Answer: (c)

1

Why: (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4.

$\frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1$ .

Q4

·1 mark·Multiple choiceRatios of standard angles

If $\tan A = \sqrt{3}$ and $\tan B = \frac{1}{\sqrt{3}}$ , where $A$ and $B$ are acute, then $A - B$ is

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Answer: (d)

30^{\circ}

Why: tan A = √3 gives A = 60°; tan B = 1/√3 gives B = 30°.

$\tan 60^{\circ} = \sqrt{3}$ and $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$ , so $A = 60^{\circ}$ , $B = 30^{\circ}$ and $A - B = 30^{\circ}$ .

Q5

·1 mark·Multiple choiceRatios of standard angles

Without a calculator, which of the following has the greatest value?

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Answer: (d)

\sec 30^{\circ}

Why: sec 30° = 2/√3, which is more than 1; the others are √3/2, 1/2 and 1.

$\sin 60^{\circ} = \frac{\sqrt{3}}{2} < 1$ , $\cos 60^{\circ} = \frac{1}{2}$ , $\tan 45^{\circ} = 1$ , $\sec 30^{\circ} = \frac{2}{\sqrt{3}}$ . Since $2 > \sqrt{3}$ , $\sec 30^{\circ} > 1$ , so it is the greatest.

Q6

·1 mark·Multiple choiceRatios of standard angles

If $\sin θ = \cos θ$ and $θ$ is acute, then $θ$ is

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Answer: (c)

45^{\circ}

Why: sin θ = cos θ means tan θ = 1.

Dividing by $\cos θ$ : $\tan θ = 1$ , so $θ = 45^{\circ}$ .

Q7

·1 mark·Multiple choiceRatios of standard angles

If $\sqrt{3} \tan θ = 3$ , where $θ$ is acute, then $\sin θ$ is

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Answer: (a)

\frac{\sqrt{3}}{2}

Why: tan θ = 3/√3 = √3, so θ = 60°.

$\tan θ = \frac{3}{\sqrt{3}} = \sqrt{3} \Rightarrow θ = 60^{\circ}$ , so $\sin θ = \frac{\sqrt{3}}{2}$ .

Q8

·1 mark·Multiple choiceIdentity sin²A + cos²A = 1 and its applications

$(1 + \tan^{2} A) \cos^{2} A$ equals

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Answer: (c)

1

Why: 1 + tan² A = sec² A and sec² A · cos² A = 1.

$(1 + \tan^{2} A) \cos^{2} A = \cos^{2} A + \sin^{2} A = 1$ .

Q9

·1 mark·Multiple choiceIdentity sin²A + cos²A = 1 and its applications

$(cosec A - \cot A) (1 + \cos A)$ equals

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Answer: (b)

\sin A

Why: Write it as (1 − cos A)(1 + cos A)/sin A = sin² A/sin A.

$\frac{1 - \cos A}{\sin A} \cdot (1 + \cos A) = \frac{1 - \cos^{2} A}{\sin A} = \frac{\sin^{2} A}{\sin A} = \sin A$ .

Q10

·1 mark·Multiple choiceTrigonometric ratios

If $\sin A = \frac{3}{5}$ , where $A$ is acute, then $\cos A + \tan A$ equals

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Answer: (c)

\frac{31}{20}

Why: cos A = 4/5 and tan A = 3/4.

$\cos A = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ , $\tan A = \frac{3 / 5}{4 / 5} = \frac{3}{4}$ . Sum $= \frac{16 + 15}{20} = \frac{31}{20}$ .

Q11

·1 mark·Multiple choiceRatios of standard angles

The value of $\sin 0^{\circ} + \cos 0^{\circ} + \tan 45^{\circ}$ is

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Answer: (d)

2

Why: 0 + 1 + 1.

$\sin 0^{\circ} = 0$ , $\cos 0^{\circ} = 1$ , $\tan 45^{\circ} = 1$ . Sum $= 2$ .

Q12

·1 mark·Multiple choiceIdentity sin²A + cos²A = 1 and its applications

If $cosec θ - \cot θ = \frac{1}{4}$ , then $cosec θ$ equals

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Answer: (a)

\frac{17}{8}

Why: cosec² θ − cot² θ = 1, so cosec θ + cot θ = 4; add the two equations.

$(cosec θ - \cot θ) (cosec θ + \cot θ) = 1 \Rightarrow cosec θ + \cot θ = 4$ . Adding: $2 cosec θ = \frac{17}{4}$ , so $cosec θ = \frac{17}{8}$ .

Q13

·1 mark·Multiple choiceRatios of standard angles

For $A = 30^{\circ}$ , which of the following is true?

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Answer: (b)

\cos 2 A = 1 - 2 \sin^{2} A

Why: cos 60° = 1/2 and 1 − 2 sin² 30° = 1 − 1/2 = 1/2.

$\sin 60^{\circ} = \frac{\sqrt{3}}{2} \neq 2 \cdot \frac{\sqrt{3}}{2}$ ; $\cos 60^{\circ} = \frac{1}{2} = 1 - 2 \cdot \frac{1}{4}$ ✓; $\tan 60^{\circ} = \sqrt{3} \neq \frac{2}{\sqrt{3}}$ ; $\cos 60^{\circ} = \frac{1}{2} \neq \sqrt{3}$ .

Q14

·1 mark·Multiple choiceIdentity sin²A + cos²A = 1 and its applications

$\sin^{4} θ - \cos^{4} θ$ equals

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Answer: (a)

1 - 2 \cos^{2} θ

Why: Difference of squares: (sin²θ − cos²θ)(sin²θ + cos²θ).

$\sin^{4} θ - \cos^{4} θ = (\sin^{2} θ - \cos^{2} θ) (\sin^{2} θ + \cos^{2} θ) = \sin^{2} θ - \cos^{2} θ = 1 - 2 \cos^{2} θ$ .

Q15

·1 mark·Multiple choiceTrigonometric ratios

If $5 \sin θ = 4$ , where $θ$ is acute, then $\sec θ + \tan θ$ equals

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Answer: (d)

3

Why: sin θ = 4/5 gives cos θ = 3/5, so sec θ = 5/3 and tan θ = 4/3.

$\sin θ = \frac{4}{5}$ , $\cos θ = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$ . $\sec θ + \tan θ = \frac{5}{3} + \frac{4}{3} = 3$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Solar panel frame (4 marks)

A solar panel on a flat roof rests on a frame shaped like a right triangle $A B C$ , right-angled at $B$ . The horizontal bar $A B$ is $12$ dm long and the vertical post $B C$ is $5$ dm tall. The panel lies along $A C$ , and $θ = ∠ B A C$ is its angle of tilt.

(i) Find the length of the panel

A C

. [1 mark]

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Answer:

13

A C = \sqrt{12^{2} + 5^{2}} = \sqrt{169} = 13

dm. A1

(ii) Write the values of

\sin θ

and

\cos θ

. [1 mark]

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Answer:

\sin θ = \frac{5}{13}

\cos θ = \frac{12}{13}

\sin θ = \frac{B C}{A C} = \frac{5}{13}

\cos θ = \frac{A B}{A C} = \frac{12}{13}

. A1

(iii) Find the value of

\tan θ + \sec θ

. [2 marks]

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Answer:

\frac{3}{2}

\tan θ = \frac{5}{12}

\sec θ = \frac{13}{12}

M1; sum

= \frac{18}{12} = \frac{3}{2}

. A1

OR Find the value of

\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}

. [2 marks]

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Answer:

\frac{119}{169}

\tan^{2} θ = \frac{25}{144}

M1;

\frac{1 - \frac{25}{144}}{1 + \frac{25}{144}} = \frac{119}{169}

. A1

Case study 2: Geometry-box set squares (4 marks)

A geometry box has two set squares. One is a right triangle with angles $30^{\circ}$ , $60^{\circ}$ and $90^{\circ}$ , whose longest side (hypotenuse) is $20$ cm. The other is a right isosceles triangle with angles $45^{\circ}$ , $45^{\circ}$ , $90^{\circ}$ , whose two equal sides are $12$ cm each.

(i) Using

\sin 30^{\circ}

, find the side of the first set square opposite the

30^{\circ}

angle. [1 mark]

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Answer:

10

Opposite side

= 20 \sin 30^{\circ} = 20 \times \frac{1}{2} = 10

cm. A1

(ii) Using

\cos 30^{\circ}

, find the third side of the first set square. [1 mark]

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Answer:

10 \sqrt{3}

Adjacent side

= 20 \cos 30^{\circ} = 20 \times \frac{\sqrt{3}}{2} = 10 \sqrt{3}

cm. A1

(iii) Show that

\frac{2 \tan 30^{\circ}}{1 + \tan^{2} 30^{\circ}} = \sin 60^{\circ}

. [2 marks]

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Answer: Both equal

\frac{\sqrt{3}}{2}

\frac{2 \cdot \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4}

= \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2} = \sin 60^{\circ}

. A1

OR Find the hypotenuse of the second set square and use it to show that

\sin 45^{\circ} = \frac{1}{\sqrt{2}}

. [2 marks]

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Answer:

12 \sqrt{2}

Hypotenuse

= \sqrt{12^{2} + 12^{2}} = 12 \sqrt{2}

cm M1;

\sin 45^{\circ} = \frac{12}{12 \sqrt{2}} = \frac{1}{\sqrt{2}}

. A1

Next steps for Introduction to Trigonometry

This free set is separate from the chapter's question bank. On the Introduction to Trigonometry chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 38 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Introduction to Trigonometry: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.