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Trigonometrical Ratios Class 9 ICSE Important Questions with Answers

The 13 questions below are Trigonometrical Ratios Class 9 ICSE important questions, each with a full step-by-step answer: extra questions beyond the textbook exercises, covering 5 topics of ICSE Class 9 Maths in the question formats used in the exam.

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Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

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Key concepts
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ICSE Class 9 Trigonometrical Ratios questions ask you to write sin⁡θ,cos⁡θ,tan⁡θ\sin\theta,\cos\theta,\tan\thetasin θ,cos θ,tan θ and their reciprocals csc⁡θ,sec⁡θ,cot⁡θ\csc\theta,\sec\theta,\cot\thetacsc θ,sec θ,cot θ from the sides of a right triangle, find all ratios given one ratio using the Pythagoras theorem, evaluate expressions and solve simple right-triangle problems. Remember sin⁡θ=opphyp\sin\theta=\dfrac{\text{opp}}{\text{hyp}}sin θ=opp/hyp, cos⁡θ=adjhyp\cos\theta=\dfrac{\text{adj}}{\text{hyp}}cos θ=adj/hyp, tan⁡θ=oppadj\tan\theta=\dfrac{\text{opp}}{\text{adj}}tan θ=opp/adj.

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

About Trigonometrical Ratios

In the ICSE Class 9 Maths chapter Trigonometrical Ratios you define the six ratios sine, cosine, tangent, cosecant, secant and cotangent of an acute angle using the sides of a right-angled triangle. You learn the reciprocal and quotient relations, find the remaining ratios when one is given, and evaluate expressions. This Selina-aligned chapter builds the foundation for standard angles and the solution of right triangles.

Sides of a right triangle: opposite, adjacent, hypotenuseThe six trigonometrical ratiosReciprocal relationsFinding all ratios from one given ratioEvaluating and simplifying trigonometric expressions

Key concepts & formulas

The three main ratios

For an acute angle θ\thetaθ in a right triangle, sin⁡θ=oppositehypotenuse\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}sin θ=opposite/hypotenuse, cos⁡θ=adjacenthypotenuse\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}cos θ=adjacent/hypotenuse, tan⁡θ=oppositeadjacent\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}tan θ=opposite/adjacent.

Reciprocal ratios

csc⁡θ=1sin⁡θ\csc\theta=\dfrac{1}{\sin\theta}csc θ=1/(sin θ), sec⁡θ=1cos⁡θ\sec\theta=\dfrac{1}{\cos\theta}sec θ=1/(cos θ), cot⁡θ=1tan⁡θ\cot\theta=\dfrac{1}{\tan\theta}cot θ=1/(tan θ).

Quotient relations

tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta}tan θ=(sin θ)/(cos θ) and cot⁡θ=cos⁡θsin⁡θ\cot\theta=\dfrac{\cos\theta}{\sin\theta}cot θ=(cos θ)/(sin θ).

Pythagoras theorem

In a right triangle, (hypotenuse)2=(opposite)2+(adjacent)2(\text{hypotenuse})^2=(\text{opposite})^2+(\text{adjacent})^2(hypotenuse)²=(opposite)²+(adjacent)²; use it to find the third side before writing the ratios.

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Trigonometrical Ratios Questions and Answers

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MCQs with answers (1 mark)

Q1MCQEasy1 mark

In a right triangle, sin⁡θ\sin\thetasin θ is equal to:

  1. (a)

    adjacenthypotenuse\dfrac{\text{adjacent}}{\text{hypotenuse}}adjacent/hypotenuse

  2. (b)

    oppositehypotenuse\dfrac{\text{opposite}}{\text{hypotenuse}}opposite/hypotenuse

  3. (c)

    oppositeadjacent\dfrac{\text{opposite}}{\text{adjacent}}opposite/adjacent

  4. (d)

    hypotenuseopposite\dfrac{\text{hypotenuse}}{\text{opposite}}hypotenuse/opposite

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Answer: (b) oppositehypotenuse\dfrac{\text{opposite}}{\text{hypotenuse}}opposite/hypotenuse.

By definition, the sine of an acute angle is the ratio of the side opposite the angle to the hypotenuse.

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Q2MCQEasy1 mark

The reciprocal of cos⁡θ\cos\thetacos θ is:

  1. (a)

    sin⁡θ\sin\thetasin θ

  2. (b)

    cot⁡θ\cot\thetacot θ

  3. (c)

    sec⁡θ\sec\thetasec θ

  4. (d)

    csc⁡θ\csc\thetacsc θ

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Answer: (c) sec⁡θ\sec\thetasec θ.

By definition sec⁡θ=1cos⁡θ\sec\theta=\dfrac{1}{\cos\theta}sec θ=1/(cos θ).

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Q3MCQModerate1 mark

If sin⁡θ=35\sin\theta=\dfrac{3}{5}sin θ=3/5, then cos⁡θ\cos\thetacos θ equals:

  1. (a)

    45\dfrac{4}{5}4/5

  2. (b)

    54\dfrac{5}{4}5/4

  3. (c)

    34\dfrac{3}{4}3/4

  4. (d)

    53\dfrac{5}{3}5/3

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Answer: (a) 45\dfrac{4}{5}4/5.

With opposite =3=3=3, hypotenuse =5=5=5, adjacent =52−32=16=4=\sqrt{5^2-3^2}=\sqrt{16}=4=√(5²-3²)=√16=4, so cos⁡θ=45\cos\theta=\dfrac{4}{5}cos θ=4/5.

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Q4MCQHOTS1 mark

The value of tan⁡θ×cot⁡θ\tan\theta\times\cot\thetatan θ × cot θ is:

  1. (a)

    000

  2. (b)

    111

  3. (c)

    sin⁡θcos⁡θ\sin\theta\cos\thetasin θ cos θ

  4. (d)

    tan⁡2θ\tan^2\thetatan²θ

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Answer: (b) 111.

Since cot⁡θ=1tan⁡θ\cot\theta=\dfrac{1}{\tan\theta}cot θ=1/(tan θ), we get tan⁡θ×cot⁡θ=tan⁡θ×1tan⁡θ=1\tan\theta\times\cot\theta=\tan\theta\times\dfrac{1}{\tan\theta}=1tan θ × cot θ=tan θ × 1/(tan θ)=1.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): For an acute angle θ\thetaθ, the value of sin⁡θ\sin\thetasin θ is always less than 111.

Reason (R): In a right triangle the hypotenuse is the longest side, so the side opposite an acute angle is shorter than the hypotenuse.

  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) Since sin⁡θ=oppositehypotenuse\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}sin θ=opposite/hypotenuse and the opposite side is shorter than the hypotenuse, sin⁡θ<1\sin\theta<1sin θ<1. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

In right triangle ABCABCABC, right-angled at BBB, AB=8 cmAB=8\text{ cm}AB=8 cm, BC=6 cmBC=6\text{ cm}BC=6 cm and AC=10 cmAC=10\text{ cm}AC=10 cm. Write sin⁡C\sin Csin C and cos⁡C\cos Ccos C.

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For angle CCC: the opposite side is AB=8 cmAB=8\text{ cm}AB=8 cm, the adjacent side is BC=6 cmBC=6\text{ cm}BC=6 cm, and the hypotenuse is AC=10 cmAC=10\text{ cm}AC=10 cm.

sin⁡C=ABAC=810=45\sin C=\dfrac{AB}{AC}=\dfrac{8}{10}=\dfrac{4}{5}sin C=AB/AC=8/10=4/5.

cos⁡C=BCAC=610=35\cos C=\dfrac{BC}{AC}=\dfrac{6}{10}=\dfrac{3}{5}cos C=BC/AC=6/10=3/5.

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Q7Very ShortModerate2 marks

If tan⁡θ=512\tan\theta=\dfrac{5}{12}tan θ=5/12, find sin⁡θ\sin\thetasin θ and cos⁡θ\cos\thetacos θ.

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Take opposite =5=5=5 and adjacent =12=12=12. Then hypotenuse =52+122=25+144=169=13=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13=√(5²+12²)=√(25+144)=√169=13.

sin⁡θ=513\sin\theta=\dfrac{5}{13}sin θ=5/13 and cos⁡θ=1213\cos\theta=\dfrac{12}{13}cos θ=12/13.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

In the right triangle shown, right-angled at BBB, AB=12 cmAB=12\text{ cm}AB=12 cm and BC=5 cmBC=5\text{ cm}BC=5 cm. Find all six trigonometrical ratios of angle AAA.

ICSE Class 9 Maths — Trigonometrical Ratios: In the right triangle shown, right-angled at B, AB=12\text{ cm} and BC=5\text{ cm}. Find all six trigonometrical ratios of angle A.
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For angle AAA: opposite side =BC=5 cm=BC=5\text{ cm}=BC=5 cm, adjacent side =AB=12 cm=AB=12\text{ cm}=AB=12 cm.

Hypotenuse AC=AB2+BC2=122+52=144+25=169=13 cmAC=\sqrt{AB^2+BC^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ cm}AC=√(AB²+BC²)=√(12²+5²)=√(144+25)=√169=13 cm.

sin⁡A=513\sin A=\dfrac{5}{13}sin A=5/13, cos⁡A=1213\cos A=\dfrac{12}{13}cos A=12/13, tan⁡A=512\tan A=\dfrac{5}{12}tan A=5/12,

csc⁡A=135\csc A=\dfrac{13}{5}csc A=13/5, sec⁡A=1312\sec A=\dfrac{13}{12}sec A=13/12, cot⁡A=125\cot A=\dfrac{12}{5}cot A=12/5.

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Q9Short AnswerModerate3 marks

If cos⁡θ=725\cos\theta=\dfrac{7}{25}cos θ=7/25, evaluate sin⁡θ−cos⁡θsin⁡θ+cos⁡θ\dfrac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}(sin θ-cos θ)/(sin θ+cos θ).

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With adjacent =7=7=7, hypotenuse =25=25=25, opposite =252−72=625−49=576=24=\sqrt{25^2-7^2}=\sqrt{625-49}=\sqrt{576}=24=√(25²-7²)=√(625-49)=√576=24.

So sin⁡θ=2425\sin\theta=\dfrac{24}{25}sin θ=24/25, cos⁡θ=725\cos\theta=\dfrac{7}{25}cos θ=7/25.

sin⁡θ−cos⁡θsin⁡θ+cos⁡θ=2425−7252425+725=17253125=1731\dfrac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\dfrac{\frac{24}{25}-\frac{7}{25}}{\frac{24}{25}+\frac{7}{25}}=\dfrac{\frac{17}{25}}{\frac{31}{25}}=\dfrac{17}{31}(sin θ-cos θ)/(sin θ+cos θ)=24/25-7/2524/25+7/25=17/2531/25=17/31.

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Q10Short AnswerHOTS3 marks

If 5sin⁡θ=35\sin\theta=35 sin θ=3, find the value of sec⁡θ+tan⁡θsec⁡θ−tan⁡θ\dfrac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}(sec θ+tan θ)/(sec θ-tan θ).

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5sin⁡θ=3⇒sin⁡θ=355\sin\theta=3\Rightarrow\sin\theta=\dfrac{3}{5}5 sin θ=3⇒ sin θ=3/5, so opposite =3=3=3, hypotenuse =5=5=5, adjacent =52−32=4=\sqrt{5^2-3^2}=4=√(5²-3²)=4.

Then cos⁡θ=45\cos\theta=\dfrac{4}{5}cos θ=4/5, sec⁡θ=54\sec\theta=\dfrac{5}{4}sec θ=5/4, tan⁡θ=34\tan\theta=\dfrac{3}{4}tan θ=3/4.

sec⁡θ+tan⁡θsec⁡θ−tan⁡θ=54+3454−34=8424=82=4\dfrac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}=\dfrac{\frac{5}{4}+\frac{3}{4}}{\frac{5}{4}-\frac{3}{4}}=\dfrac{\frac{8}{4}}{\frac{2}{4}}=\dfrac{8}{2}=4(sec θ+tan θ)/(sec θ-tan θ)=5/4+3/45/4-3/4=8/42/4=8/2=4.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

In right triangle PQRPQRPQR, right-angled at QQQ, PQ=24 cmPQ=24\text{ cm}PQ=24 cm and PR=25 cmPR=25\text{ cm}PR=25 cm. (i) Find QRQRQR. (ii) Write all six trigonometrical ratios of angle RRR. (iii) Verify that tan⁡R=sin⁡Rcos⁡R\tan R=\dfrac{\sin R}{\cos R}tan R=(sin R)/(cos R).

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(i) By Pythagoras, QR=PR2−PQ2=252−242=625−576=49=7 cmQR=\sqrt{PR^2-PQ^2}=\sqrt{25^2-24^2}=\sqrt{625-576}=\sqrt{49}=7\text{ cm}QR=√(PR²-PQ²)=√(25²-24²)=√(625-576)=√49=7 cm.

(ii) For angle RRR: opposite =PQ=24=PQ=24=PQ=24, adjacent =QR=7=QR=7=QR=7, hypotenuse =PR=25=PR=25=PR=25.

sin⁡R=2425\sin R=\dfrac{24}{25}sin R=24/25, cos⁡R=725\cos R=\dfrac{7}{25}cos R=7/25, tan⁡R=247\tan R=\dfrac{24}{7}tan R=24/7,

csc⁡R=2524\csc R=\dfrac{25}{24}csc R=25/24, sec⁡R=257\sec R=\dfrac{25}{7}sec R=25/7, cot⁡R=724\cot R=\dfrac{7}{24}cot R=7/24.

(iii) sin⁡Rcos⁡R=24/257/25=247=tan⁡R\dfrac{\sin R}{\cos R}=\dfrac{24/25}{7/25}=\dfrac{24}{7}=\tan R(sin R)/(cos R)=24/25/7/25=24/7=tan R. Verified.

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Q12Long AnswerHOTS5 marks

If cot⁡θ=158\cot\theta=\dfrac{15}{8}cot θ=15/8, find the values of (i) sin⁡θ\sin\thetasin θ and cos⁡θ\cos\thetacos θ, and (ii) evaluate (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}(1+sin θ)(1-sin θ)/(1+cos θ)(1-cos θ).

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(i) cot⁡θ=adjacentopposite=158\cot\theta=\dfrac{\text{adjacent}}{\text{opposite}}=\dfrac{15}{8}cot θ=adjacent/opposite=15/8, so adjacent =15=15=15, opposite =8=8=8.

Hypotenuse =152+82=225+64=289=17=\sqrt{15^2+8^2}=\sqrt{225+64}=\sqrt{289}=17=√(15²+8²)=√(225+64)=√289=17.

sin⁡θ=817\sin\theta=\dfrac{8}{17}sin θ=8/17, cos⁡θ=1517\cos\theta=\dfrac{15}{17}cos θ=15/17.

(ii) (1+sin⁡θ)(1−sin⁡θ)=1−sin⁡2θ=1−(817)2=1−64289=225289(1+\sin\theta)(1-\sin\theta)=1-\sin^2\theta=1-\left(\dfrac{8}{17}\right)^2=1-\dfrac{64}{289}=\dfrac{225}{289}(1+sin θ)(1-sin θ)=1-sin²θ=1-(8/17)²=1-64/289=225/289.

(1+cos⁡θ)(1−cos⁡θ)=1−cos⁡2θ=1−(1517)2=1−225289=64289(1+\cos\theta)(1-\cos\theta)=1-\cos^2\theta=1-\left(\dfrac{15}{17}\right)^2=1-\dfrac{225}{289}=\dfrac{64}{289}(1+cos θ)(1-cos θ)=1-cos²θ=1-(15/17)²=1-225/289=64/289.

Ratio =225/28964/289=22564=\dfrac{225/289}{64/289}=\dfrac{225}{64}=225/289/64/289=225/64.

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Case study questions (4 marks)

Q13Case-basedModerate4 marks

A ladder leans against a wall forming a right-angled triangle with the ground. The foot of the ladder is 9 m9\text{ m}9 m from the wall (adjacent side) and the ladder reaches 12 m12\text{ m}12 m up the wall (opposite side). Let θ\thetaθ be the angle the ladder makes with the ground.

(i) Find the length of the ladder (the hypotenuse).

(ii) Write sin⁡θ\sin\thetasin θ and cos⁡θ\cos\thetacos θ.

(iii) Find tan⁡θ\tan\thetatan θ.

(iv) Show that sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin²θ+cos²θ=1.

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(i) Ladder length =92+122=81+144=225=15 m=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15\text{ m}=√(9²+12²)=√(81+144)=√225=15 m.

(ii) sin⁡θ=oppositehypotenuse=1215=45\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{12}{15}=\dfrac{4}{5}sin θ=opposite/hypotenuse=12/15=4/5; cos⁡θ=adjacenthypotenuse=915=35\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{9}{15}=\dfrac{3}{5}cos θ=adjacent/hypotenuse=9/15=3/5.

(iii) tan⁡θ=oppositeadjacent=129=43\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{12}{9}=\dfrac{4}{3}tan θ=opposite/adjacent=12/9=4/3.

(iv) sin⁡2θ+cos⁡2θ=(45)2+(35)2=1625+925=2525=1\sin^2\theta+\cos^2\theta=\left(\dfrac{4}{5}\right)^2+\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}+\dfrac{9}{25}=\dfrac{25}{25}=1sin²θ+cos²θ=(4/5)²+(3/5)²=16/25+9/25=25/25=1.

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Continues in Class 10: Trigonometrical Identities 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 six trigonometric ratios in ICSE Class 9?
    sin⁡θ=opphyp\sin\theta=\tfrac{\text{opp}}{\text{hyp}}sin θ=opp/hyp, cos⁡θ=adjhyp\cos\theta=\tfrac{\text{adj}}{\text{hyp}}cos θ=adj/hyp, tan⁡θ=oppadj\tan\theta=\tfrac{\text{opp}}{\text{adj}}tan θ=opp/adj, and their reciprocals csc⁡θ\csc\thetacsc θ, sec⁡θ\sec\thetasec θ and cot⁡θ\cot\thetacot θ. Also tan⁡θ=sin⁡θcos⁡θ\tan\theta=\tfrac{\sin\theta}{\cos\theta}tan θ=(sin θ)/(cos θ). This page has 13 questions using them, with answers. Learn the three main ratios well; the other three are just their reciprocals.
  • If tan θ = 5/12, how do you find sin θ and cos θ?
    Draw a right triangle with opposite side 555 and adjacent side 121212; by Pythagoras the hypotenuse is 131313. So sin⁡θ=513\sin\theta=\tfrac5{13}sin θ=5/13 and cos⁡θ=1213\cos\theta=\tfrac{12}{13}cos θ=12/13. The same method works whichever ratio is given.
  • If 5 sin θ = 3, what is (sec θ + tan θ)/(sec θ − tan θ)?
    sin⁡θ=35\sin\theta=\tfrac35sin θ=3/5, so the triangle is 3,4,53, 4, 53, 4, 5: sec⁡θ=54\sec\theta=\tfrac54sec θ=5/4 and tan⁡θ=34\tan\theta=\tfrac34tan θ=3/4. The expression is 21/2=4\dfrac{2}{1/2}=42/1/2=4. Find each ratio first, then substitute. Simplifying the numerator and denominator separately before dividing keeps the fractions manageable.
  • Why is sin θ always less than 1 for an acute angle?
    Because sin⁡θ\sin\thetasin θ is the opposite side divided by the hypotenuse, and the hypotenuse is the longest side of a right triangle. The same is true for cos⁡θ\cos\thetacos θ. tan⁡θ\tan\thetatan θ, however, can be more than 111.
  • Which ICSE Class 9 trigonometry chapters follow Trigonometrical Ratios?
    Trigonometrical Ratios of Standard Angles (Chapter 23), Solution of Right Triangles (Chapter 24) and Complementary Angles (Chapter 25). Trigonometrical Ratios is Chapter 22; each later chapter has its own important-questions page.
  • How do I avoid mixing up opposite and adjacent sides?
    Mark the angle θ\thetaθ first, label the side facing it 'opposite', the side touching it (not the hypotenuse) 'adjacent', and the longest side 'hypotenuse'. Relabel every time the angle changes, since opposite and adjacent swap.
  • Can I download Trigonometrical Ratios ICSE Class 9 questions?
    Yes. Enter your email in the free PDF box on this page and all 13 ratio questions, the ladder case study included, open with answers as a printable worksheet.

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