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Chapter 23: Trigonometrical Ratios of Standard AnglesICSE Class 9 Maths100% Free

Trig Ratios of Standard Angles Class 9 ICSE Important Questions with Answers

The 13 questions below are Trig Ratios of Standard Angles Class 9 ICSE important questions, each with a full step-by-step answer: extra questions beyond the textbook exercises, covering 6 topics of ICSE Class 9 Maths in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

13
Questions
6
Topics
4
Key concepts
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With answers
Quick answer

ICSE Class 9 standard-angle questions ask you to recall the exact values of sin⁡,cos⁡,tan⁡\sin,\cos,\tansin ,cos ,tan at 0∘,30∘,45∘,60∘,90∘0^\circ,30^\circ,45^\circ,60^\circ,90^\circ0°,30°,45°,60°,90° (for example sin⁡30∘=12\sin30^\circ=\dfrac{1}{2}sin 30°=1/2, tan⁡45∘=1\tan45^\circ=1tan 45°=1), use them to evaluate expressions and verify identities, and solve one right-angled triangle, such as the height of a kite or the shadow of a tower.

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Trigonometrical Ratios of Standard Angles

In this 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 equationSimple right-triangle problems

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=1sin 0°=0, sin 30°=1/2, sin 45°=1/√2, sin 60°=√3/2, sin 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=\sqrt3tan 0°=0, tan 30°=1/√3, tan 45°=1, tan 60°=√3, and tan⁡90∘\tan90^\circtan 90° 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=\sqrt2csc 30°=2, sec 60°=2, cot 45°=1, sec 45°=csc 45°=√2.

Useful checks

sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin²θ+cos²θ=1 holds at every standard angle; e.g. sin⁡230∘+cos⁡230∘=14+34=1\sin^230^\circ+\cos^230^\circ=\dfrac14+\dfrac34=1sin²30°+cos²30°=1/4+3/4=1.

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Trig Ratios of Standard Angles Questions and Answers

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MCQs with answers (1 mark)

Q1MCQEasy1 mark

The value of sin⁡30∘\sin30^\circsin 30° 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}sin 30°=1/2.

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Q2MCQEasy1 mark

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

  1. (a)

    000

  2. (b)

    111

  3. (c)

    3\sqrt3√3

  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}=1tan 45°=(sin 45°)/(cos 45°)=1/√2/1/√2=1.

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Q3MCQModerate1 mark

The value of 2sin⁡30∘cos⁡30∘2\sin30^\circ\cos30^\circ2 sin 30° cos 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\sqrt3√3

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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}2 sin 30° cos 30°=2× 1/2× √3/2=√3/2.

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Q4MCQHOTS1 mark

If sin⁡(A)=cos⁡(A)\sin(A)=\cos(A)sin (A)=cos (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}sin 45°=cos 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^\circtan 90° is not defined.

Reason (R): tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta}tan θ=(sin θ)/(cos θ) and cos⁡90∘=0\cos90^\circ=0cos 90°=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}tan 90°=(sin 90°)/(cos 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^\circsin²30°+cos²60°.

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sin⁡30∘=12\sin30^\circ=\dfrac{1}{2}sin 30°=1/2 and cos⁡60∘=12\cos60^\circ=\dfrac{1}{2}cos 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}sin²30°+cos²60°=(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}tan 60°-1/(tan 30°).

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tan⁡60∘=3\tan60^\circ=\sqrt3tan 60°=√3 and tan⁡30∘=13\tan30^\circ=\dfrac{1}{\sqrt3}tan 30°=1/√3, so 1tan⁡30∘=3\dfrac{1}{\tan30^\circ}=\sqrt31/(tan 30°)=√3.

tan⁡60∘−1tan⁡30∘=3−3=0\tan60^\circ-\dfrac{1}{\tan30^\circ}=\sqrt3-\sqrt3=0tan 60°-1/(tan 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}(sin 30°+cos 60°)/(tan 45°).

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

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Q9Short AnswerModerate3 marks

A kite is flying on a string 100100100 m long. The string is straight and makes an angle of 30∘30^\circ30° with the level ground. Find the height of the kite.

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The string is the hypotenuse of a right triangle and the height hhh is the side opposite the 30∘30^\circ30° angle:
sin⁡30∘=h100⇒h=100×12=50 m.\sin30^\circ=\frac{h}{100}\Rightarrow h=100\times\frac12=50\ \text{m}.sin 30°=h/100⇒ h=100× 1/2=50 m.

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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 cot²45°-sec²60°+sin²60°+cos²90°.

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

4cot⁡245∘=4(1)2=44\cot^245^\circ=4(1)^2=44 cot²45°=4(1)²=4.

sec⁡260∘=22=4\sec^260^\circ=2^2=4sec²60°=2²=4.

sin⁡260∘=(32)2=34\sin^260^\circ=\left(\dfrac{\sqrt3}{2}\right)^2=\dfrac{3}{4}sin²60°=(√3/2)²=3/4.

cos⁡290∘=0\cos^290^\circ=0cos²90°=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}(tan²60°+4 cos²45°+3 sec²30°+5 cos²90°)/(csc 30°+sec 60°-cot²30°).

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Standard values: tan⁡60∘=3\tan60^\circ=\sqrt3tan 60°=√3, cos⁡45∘=12\cos45^\circ=\dfrac{1}{\sqrt2}cos 45°=1/√2, sec⁡30∘=23\sec30^\circ=\dfrac{2}{\sqrt3}sec 30°=2/√3, cos⁡90∘=0\cos90^\circ=0cos 90°=0, csc⁡30∘=2\csc30^\circ=2csc 30°=2, sec⁡60∘=2\sec60^\circ=2sec 60°=2, cot⁡30∘=3\cot30^\circ=\sqrt3cot 30°=√3.

Numerator:
tan⁡260∘=3\tan^260^\circ=3tan²60°=3; 4cos⁡245∘=4×12=24\cos^245^\circ=4\times\dfrac{1}{2}=24 cos²45°=4× 1/2=2; 3sec⁡230∘=3×43=43\sec^230^\circ=3\times\dfrac{4}{3}=43 sec²30°=3× 4/3=4; 5cos⁡290∘=05\cos^290^\circ=05 cos²90°=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=1csc 30°+sec 60°-cot²30°=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^\circθ=30°, verify that (i) sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\thetasin 2θ=2 sin θ cos θ and (ii) cos⁡2θ=1−2sin⁡2θ\cos2\theta=1-2\sin^2\thetacos 2θ=1-2 sin²θ.

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Here θ=30∘\theta=30^\circθ=30°, so 2θ=60∘2\theta=60^\circ2θ=60°. Use sin⁡30∘=12\sin30^\circ=\dfrac12sin 30°=1/2, cos⁡30∘=32\cos30^\circ=\dfrac{\sqrt3}{2}cos 30°=√3/2, sin⁡60∘=32\sin60^\circ=\dfrac{\sqrt3}{2}sin 60°=√3/2, cos⁡60∘=12\cos60^\circ=\dfrac12cos 60°=1/2.

(i) Left side =sin⁡60∘=32=\sin60^\circ=\dfrac{\sqrt3}{2}=sin 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}=2 sin 30° cos 30°=2× 1/2× √3/2=√3/2.

Left === Right, verified.

(ii) Left side =cos⁡60∘=12=\cos60^\circ=\dfrac{1}{2}=cos 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 sin²30°=1-2(1/2)²=1-1/2=1/2.

Left === Right, verified.

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Case study questions (4 marks)

Q13Case-basedHOTS4 marks

Read the passage and answer the questions.

A vertical tower is 303030 m tall. In the morning the Sun's rays make an angle of 30∘30^\circ30° with the ground; at noon the angle is 60∘60^\circ60°. The tower, its shadow and the Sun's ray form a right-angled triangle.

(i) State the value of tan⁡60∘\tan60^\circtan 60°.

(ii) Find the length of the shadow at noon.

(iii) Find the length of the shadow in the morning.

(iv) By how much is the morning shadow longer? (Take 3=1.732\sqrt3=1.732√3=1.732.)

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(i) tan⁡60∘=3\tan60^\circ=\sqrt3tan 60°=√3.

(ii) tan⁡60∘=30s⇒s=303=103≈17.32\tan60^\circ=\dfrac{30}{s}\Rightarrow s=\dfrac{30}{\sqrt3}=10\sqrt3\approx17.32tan 60°=30/s⇒ s=30/√3=10√3≈17.32 m.

(iii) tan⁡30∘=30s⇒s=301/3=303≈51.96\tan30^\circ=\dfrac{30}{s}\Rightarrow s=\dfrac{30}{1/\sqrt3}=30\sqrt3\approx51.96tan 30°=30/s⇒ s=30/1/√3=30√3≈51.96 m.

(iv) 303−103=203≈34.6430\sqrt3-10\sqrt3=20\sqrt3\approx34.6430√3-10√3=20√3≈34.64 m longer.

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Continues in Class 10: Heights and Distances Class 10 ICSE questions

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

Frequently asked questions

  • What are the trigonometric values of standard angles?
    sin⁡\sinsin of 0∘,30∘,45∘,60∘,90∘0^\circ,30^\circ,45^\circ,60^\circ,90^\circ0°,30°,45°,60°,90° is 0,12,12,32,10,\tfrac12,\tfrac1{\sqrt2},\tfrac{\sqrt3}{2},10,1/2,1/√2,√3/2,1, and cosine runs in reverse. tan⁡\tantan is 0,13,1,30,\tfrac1{\sqrt3},1,\sqrt30,1/√3,1,√3 and is undefined at 90∘90^\circ90°. Reciprocals follow: csc⁡30∘=2\csc30^\circ=2csc 30°=2, sec⁡60∘=2\sec60^\circ=2sec 60°=2. These values appear in almost every ICSE trigonometry question from here to Class 10.
  • If sin A = cos A, what is A?
    A=45∘A=45^\circA=45°, the only acute angle where the opposite and adjacent sides are equal, giving sin⁡45∘=cos⁡45∘=12\sin45^\circ=\cos45^\circ=\tfrac1{\sqrt2}sin 45°=cos 45°=1/√2. Equivalently, tan⁡A=1\tan A=1tan A=1. This idea of matching sine and cosine leads straight into the Complementary Angles chapter, where sinA=cosBsin A=cos Bsin A=cos B means the two angles add up to 90circ90^circ90^circ.
  • How do you evaluate 4cot²45° − sec²60° + sin²60° + cos²90°?
    Substitute: 4(1)2−(2)2+(32)2+02=4−4+34+0=344(1)^2-(2)^2+\left(\tfrac{\sqrt3}{2}\right)^2+0^2=4-4+\tfrac34+0=\tfrac344(1)²-(2)²+(√3/2)²+0²=4-4+3/4+0=3/4. Square each value carefully, since sec⁡260∘\sec^260^\circsec²60° means (sec⁡60∘)2(\sec60^\circ)^2(sec 60°)². Write every standard value on one line before you start, then square and combine them one term at a time. A single wrong value spoils the whole answer, so check each against the table.
  • How high is a kite on a 100 m string at 30° to the ground?
    The string is the hypotenuse, so height =100sin⁡30∘=100×12=50=100\sin30^\circ=100\times\tfrac12=50=100 sin 30°=100× 1/2=50 m. Standard angles give exact answers, so no tables are needed. If the angle were 60circ60^circ60^circ, the height would be 50sqrt350sqrt350sqrt3 m instead.
  • Which ICSE Class 9 chapter uses standard angle values next?
    Solution of Right Triangles (Chapter 24) uses them to find unknown sides and angles. Standard Angles is Chapter 23, after Trigonometrical Ratios (Chapter 22). Both neighbours have important-questions pages.
  • What is the quickest way to learn the standard angle table?
    For sine, take 02,12,22,32,42\tfrac{\sqrt0}{2},\tfrac{\sqrt1}{2},\tfrac{\sqrt2}{2},\tfrac{\sqrt3}{2},\tfrac{\sqrt4}{2}√0/2,√1/2,√2/2,√3/2,√4/2; reverse it for cosine and divide for tangent. Check any row with sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin²θ+cos²θ=1, which holds at every angle. Ten minutes of writing it from memory each day fixes it quickly.
  • Can I download standard angle trigonometry questions for ICSE Class 9?
    Yes. Enter your email in the free PDF box on this page and all 13 standard-angle questions open with full substitutions as a printable worksheet.

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