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Trigonometrical Ratios of Standard Angles — ICSE Class 9 Maths Important Questions

13 hand-picked ICSE Class 9 Maths important questions for Trigonometrical Ratios of Standard Angles, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
Questions
6
Question types
32
Total marks
₹0
With answers
Quick answer

High-yield ICSE Class 9 questions on standard angles ask you to recall the exact values of sin⁡,cos⁡,tan⁡\sin,\cos,\tan,, at 0∘,30∘,45∘,60∘,90∘0^\circ,30^\circ,45^\circ,60^\circ,90^\circ0^,30^,45^,60^,90^ and use them to evaluate or simplify expressions and to verify identities. Key values: sin⁡30∘=12\sin30^\circ=\dfrac{1}{2}30^=1/2, cos⁡30∘=32\cos30^\circ=\dfrac{\sqrt3}{2}30^=3/2, tan⁡45∘=1\tan45^\circ=145^=1, sin⁡60∘=32\sin60^\circ=\dfrac{\sqrt3}{2}60^=3/2.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Trigonometrical Ratios of Standard Angles

In the ICSE Class 9 Maths chapter Trigonometrical Ratios of Standard Angles you memorise and apply the exact values of the trigonometric ratios at 0∘,30∘,45∘,60∘0^\circ,30^\circ,45^\circ,60^\circ0^,30^,45^,60^ and 90∘90^\circ90^. Typical Selina-aligned questions ask you to evaluate numerical expressions, simplify combinations of these ratios, verify identities, and find an unknown angle from an equation. Accurate recall of the standard-angle table is the core skill.

The standard-angle value tableEvaluating numerical expressionsSimplifying combinations of ratiosVerifying identities at standard anglesFinding an angle from an equation

Key concepts & formulas

Standard-angle table

sin⁡0∘=0, sin⁡30∘=12, sin⁡45∘=12, sin⁡60∘=32, sin⁡90∘=1\sin0^\circ=0,\ \sin30^\circ=\dfrac12,\ \sin45^\circ=\dfrac{1}{\sqrt2},\ \sin60^\circ=\dfrac{\sqrt3}{2},\ \sin90^\circ=10^=0, 30^=12, 45^=1/2, 60^=3/2, 90^=1. The cosine values run in the reverse order.

Tangent values

tan⁡0∘=0, tan⁡30∘=13, tan⁡45∘=1, tan⁡60∘=3\tan0^\circ=0,\ \tan30^\circ=\dfrac{1}{\sqrt3},\ \tan45^\circ=1,\ \tan60^\circ=\sqrt30^=0, 30^=1/3, 45^=1, 60^=3, and tan⁡90∘\tan90^\circ90^ is not defined.

Reciprocals at standard angles

csc⁡30∘=2, sec⁡60∘=2, cot⁡45∘=1, sec⁡45∘=csc⁡45∘=2\csc30^\circ=2,\ \sec60^\circ=2,\ \cot45^\circ=1,\ \sec45^\circ=\csc45^\circ=\sqrt230^=2, 60^=2, 45^=1, 45^=45^=2.

Useful checks

sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1^2+^2=1 holds at every standard angle; e.g. sin⁡230∘+cos⁡230∘=14+34=1\sin^230^\circ+\cos^230^\circ=\dfrac14+\dfrac34=1^230^+^230^=14+34=1.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The value of sin⁡30∘\sin30^\circ30^ is:

  1. (a)

    12\dfrac{1}{2}1/2

  2. (b)

    32\dfrac{\sqrt3}{2}3/2

  3. (c)

    111

  4. (d)

    12\dfrac{1}{\sqrt2}1/2

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Answer: (a) 12\dfrac{1}{2}1/2.

From the standard-angle table, sin⁡30∘=12\sin30^\circ=\dfrac{1}{2}30^=1/2.

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Q2MCQEasy1 mark

The value of tan⁡45∘\tan45^\circ45^ is:

  1. (a)

    000

  2. (b)

    111

  3. (c)

    3\sqrt33

  4. (d)

    13\dfrac{1}{\sqrt3}1/3

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Answer: (b) 111.

tan⁡45∘=sin⁡45∘cos⁡45∘=1/21/2=1\tan45^\circ=\dfrac{\sin45^\circ}{\cos45^\circ}=\dfrac{1/\sqrt2}{1/\sqrt2}=145^=45^/45^=1/2/1/2=1.

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Q3MCQModerate1 mark

The value of 2sin⁡30∘cos⁡30∘2\sin30^\circ\cos30^\circ230^30^ is:

  1. (a)

    12\dfrac{1}{2}1/2

  2. (b)

    32\dfrac{\sqrt3}{2}3/2

  3. (c)

    13\dfrac{1}{\sqrt3}1/3

  4. (d)

    3\sqrt33

Show model answer

Answer: (b) 32\dfrac{\sqrt3}{2}3/2.

2sin⁡30∘cos⁡30∘=2×12×32=322\sin30^\circ\cos30^\circ=2\times\dfrac{1}{2}\times\dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{2}230^30^=2×1/2×3/2=3/2.

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Q4MCQHOTS1 mark

If sin⁡(A)=cos⁡(A)\sin(A)=\cos(A)(A)=(A) for an acute angle AAA, then AAA equals:

  1. (a)

    30∘30^\circ30^

  2. (b)

    45∘45^\circ45^

  3. (c)

    60∘60^\circ60^

  4. (d)

    90∘90^\circ90^

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Answer: (b) 45∘45^\circ45^.

Since sin⁡45∘=cos⁡45∘=12\sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}45^=45^=1/2, the acute angle for which sine equals cosine is 45∘45^\circ45^.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): tan⁡90∘\tan90^\circ90^ is not defined.

Reason (R): tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta}=/ and cos⁡90∘=0\cos90^\circ=090^=0.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) tan⁡90∘=sin⁡90∘cos⁡90∘=10\tan90^\circ=\dfrac{\sin90^\circ}{\cos90^\circ}=\dfrac{1}{0}90^=90^/90^=1/0, which is not defined because division by 000 is undefined. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Evaluate sin⁡230∘+cos⁡260∘\sin^230^\circ+\cos^260^\circ^230^+^260^.

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sin⁡30∘=12\sin30^\circ=\dfrac{1}{2}30^=1/2 and cos⁡60∘=12\cos60^\circ=\dfrac{1}{2}60^=1/2.

sin⁡230∘+cos⁡260∘=(12)2+(12)2=14+14=12\sin^230^\circ+\cos^260^\circ=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}^230^+^260^=(1/2)^2+(1/2)^2=1/4+1/4=1/2.

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Q7Very ShortModerate2 marks

Evaluate tan⁡60∘−1tan⁡30∘\tan60^\circ-\dfrac{1}{\tan30^\circ}60^-1/30^.

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tan⁡60∘=3\tan60^\circ=\sqrt360^=3 and tan⁡30∘=13\tan30^\circ=\dfrac{1}{\sqrt3}30^=1/3, so 1tan⁡30∘=3\dfrac{1}{\tan30^\circ}=\sqrt31/30^=3.

tan⁡60∘−1tan⁡30∘=3−3=0\tan60^\circ-\dfrac{1}{\tan30^\circ}=\sqrt3-\sqrt3=060^-1/30^=3-3=0.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Evaluate sin⁡30∘+cos⁡60∘tan⁡45∘\dfrac{\sin30^\circ+\cos60^\circ}{\tan45^\circ}30^+60^/45^.

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sin⁡30∘=12\sin30^\circ=\dfrac{1}{2}30^=1/2, cos⁡60∘=12\cos60^\circ=\dfrac{1}{2}60^=1/2, tan⁡45∘=1\tan45^\circ=145^=1.

Numerator =12+12=1=\dfrac{1}{2}+\dfrac{1}{2}=1=1/2+1/2=1.

sin⁡30∘+cos⁡60∘tan⁡45∘=11=1\dfrac{\sin30^\circ+\cos60^\circ}{\tan45^\circ}=\dfrac{1}{1}=130^+60^/45^=1/1=1.

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Q9Short AnswerModerate3 marks

Show that cos⁡60∘=1−2sin⁡230∘\cos60^\circ=1-2\sin^230^\circ60^=1-2^230^.

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Left side: cos⁡60∘=12\cos60^\circ=\dfrac{1}{2}60^=1/2.

Right side: 1−2sin⁡230∘=1−2(12)2=1−2×14=1−12=121-2\sin^230^\circ=1-2\left(\dfrac{1}{2}\right)^2=1-2\times\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}1-2^230^=1-2(1/2)^2=1-2×1/4=1-1/2=1/2.

Since left side === right side =12=\dfrac{1}{2}=1/2, the result is verified.

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Q10Short AnswerHOTS3 marks

Evaluate 4cot⁡245∘−sec⁡260∘+sin⁡260∘+cos⁡290∘4\cot^245^\circ-\sec^260^\circ+\sin^260^\circ+\cos^290^\circ4^245^-^260^+^260^+^290^.

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Values: cot⁡45∘=1\cot45^\circ=145^=1, sec⁡60∘=2\sec60^\circ=260^=2, sin⁡60∘=32\sin60^\circ=\dfrac{\sqrt3}{2}60^=3/2, cos⁡90∘=0\cos90^\circ=090^=0.

4cot⁡245∘=4(1)2=44\cot^245^\circ=4(1)^2=44^245^=4(1)^2=4.

sec⁡260∘=22=4\sec^260^\circ=2^2=4^260^=2^2=4.

sin⁡260∘=(32)2=34\sin^260^\circ=\left(\dfrac{\sqrt3}{2}\right)^2=\dfrac{3}{4}^260^=(3/2)^2=3/4.

cos⁡290∘=0\cos^290^\circ=0^290^=0.

Sum =4−4+34+0=34=4-4+\dfrac{3}{4}+0=\dfrac{3}{4}=4-4+3/4+0=3/4.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Evaluate tan⁡260∘+4cos⁡245∘+3sec⁡230∘+5cos⁡290∘csc⁡30∘+sec⁡60∘−cot⁡230∘\dfrac{\tan^260^\circ+4\cos^245^\circ+3\sec^230^\circ+5\cos^290^\circ}{\csc30^\circ+\sec60^\circ-\cot^230^\circ}^260^+4^245^+3^230^+5^290^/30^+60^-^230^.

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Standard values: tan⁡60∘=3\tan60^\circ=\sqrt360^=3, cos⁡45∘=12\cos45^\circ=\dfrac{1}{\sqrt2}45^=1/2, sec⁡30∘=23\sec30^\circ=\dfrac{2}{\sqrt3}30^=2/3, cos⁡90∘=0\cos90^\circ=090^=0, csc⁡30∘=2\csc30^\circ=230^=2, sec⁡60∘=2\sec60^\circ=260^=2, cot⁡30∘=3\cot30^\circ=\sqrt330^=3.

Numerator:
tan⁡260∘=3\tan^260^\circ=3^260^=3; 4cos⁡245∘=4×12=24\cos^245^\circ=4\times\dfrac{1}{2}=24^245^=4×1/2=2; 3sec⁡230∘=3×43=43\sec^230^\circ=3\times\dfrac{4}{3}=43^230^=3×4/3=4; 5cos⁡290∘=05\cos^290^\circ=05^290^=0.

Numerator =3+2+4+0=9=3+2+4+0=9=3+2+4+0=9.

Denominator:
csc⁡30∘+sec⁡60∘−cot⁡230∘=2+2−(3)2=2+2−3=1\csc30^\circ+\sec60^\circ-\cot^230^\circ=2+2-(\sqrt3)^2=2+2-3=130^+60^-^230^=2+2-(3)^2=2+2-3=1.

Value =91=9=\dfrac{9}{1}=9=9/1=9.

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Q12Long AnswerHOTS5 marks

If θ=30∘\theta=30^\circ=30^, verify that (i) sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta2=2 and (ii) cos⁡2θ=1−2sin⁡2θ\cos2\theta=1-2\sin^2\theta2=1-2^2.

Show model answer

Here θ=30∘\theta=30^\circ=30^, so 2θ=60∘2\theta=60^\circ2=60^. Use sin⁡30∘=12\sin30^\circ=\dfrac1230^=12, cos⁡30∘=32\cos30^\circ=\dfrac{\sqrt3}{2}30^=3/2, sin⁡60∘=32\sin60^\circ=\dfrac{\sqrt3}{2}60^=3/2, cos⁡60∘=12\cos60^\circ=\dfrac1260^=12.

(i) Left side =sin⁡60∘=32=\sin60^\circ=\dfrac{\sqrt3}{2}=60^=3/2.

Right side =2sin⁡30∘cos⁡30∘=2×12×32=32=2\sin30^\circ\cos30^\circ=2\times\dfrac{1}{2}\times\dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{2}=230^30^=2×1/2×3/2=3/2.

Left === Right, verified.

(ii) Left side =cos⁡60∘=12=\cos60^\circ=\dfrac{1}{2}=60^=1/2.

Right side =1−2sin⁡230∘=1−2(12)2=1−12=12=1-2\sin^230^\circ=1-2\left(\dfrac{1}{2}\right)^2=1-\dfrac{1}{2}=\dfrac{1}{2}=1-2^230^=1-2(1/2)^2=1-1/2=1/2.

Left === Right, verified.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A student is asked to evaluate an expression built from standard angles. Using the standard-angle table, answer the following.

(i) State the value of cos⁡30∘\cos30^\circ30^.

(ii) Evaluate sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ60^30^+60^30^.

(iii) Which single standard angle has this value as its sine?

(iv) Hence write the value of the given expression as sin⁡( ? )\sin(\,?\,)(\,?\,).

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(i) cos⁡30∘=32\cos30^\circ=\dfrac{\sqrt3}{2}30^=3/2.

(ii) sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘=32×32+12×12=34+14=1\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ=\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}+\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{3}{4}+\dfrac{1}{4}=160^30^+60^30^=3/2×3/2+1/2×1/2=3/4+1/4=1.

(iii) The value 111 is the sine of 90∘90^\circ90^, since sin⁡90∘=1\sin90^\circ=190^=1.

(iv) The expression equals sin⁡(90∘)\sin(90^\circ)(90^), consistent with sin⁡(60∘+30∘)=sin⁡90∘=1\sin(60^\circ+30^\circ)=\sin90^\circ=1(60^+30^)=90^=1.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

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    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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