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Trigonometrical Ratios — ICSE Class 9 Maths Important Questions

13 hand-picked ICSE Class 9 Maths important questions for Trigonometrical Ratios, 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
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Quick answer

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

Syllabus reference (2026–27): 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\theta in a right triangle, sin⁡θ=oppositehypotenuse\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=opposite/hypotenuse, cos⁡θ=adjacenthypotenuse\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=adjacent/hypotenuse, tan⁡θ=oppositeadjacent\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=opposite/adjacent.

Reciprocal ratios

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

Quotient relations

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

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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Important questions with answers

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

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

In a right triangle, sin⁡θ\sin\theta 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\theta is:

  1. (a)

    sin⁡θ\sin\theta

  2. (b)

    cot⁡θ\cot\theta

  3. (c)

    sec⁡θ\sec\theta

  4. (d)

    csc⁡θ\csc\theta

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Answer: (c) sec⁡θ\sec\theta.

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

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Q3MCQModerate1 mark

If sin⁡θ=35\sin\theta=\dfrac{3}{5}=3/5, then cos⁡θ\cos\theta 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=√5^2-3^2=√16=4, so cos⁡θ=45\cos\theta=\dfrac{4}{5}=4/5.

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Q4MCQHOTS1 mark

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

  1. (a)

    000

  2. (b)

    111

  3. (c)

    sin⁡θcos⁡θ\sin\theta\cos\theta

  4. (d)

    tan⁡2θ\tan^2\theta^2

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Answer: (b) 111.

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

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): For an acute angle θ\theta, the value of sin⁡θ\sin\theta 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}}=opposite/hypotenuse and the opposite side is shorter than the hypotenuse, sin⁡θ<1\sin\theta<1<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 CC and cos⁡C\cos CC.

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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}C=AB/AC=8/10=4/5.

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

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Q7Very ShortModerate2 marks

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

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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=√5^2+12^2=√25+144=√169=13.

sin⁡θ=513\sin\theta=\dfrac{5}{13}=5/13 and cos⁡θ=1213\cos\theta=\dfrac{12}{13}=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=√AB^2+BC^2=√12^2+5^2=√144+25=√169=13 cm.

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

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

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Q9Short AnswerModerate3 marks

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

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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=√25^2-7^2=√625-49=√576=24.

So sin⁡θ=2425\sin\theta=\dfrac{24}{25}=24/25, cos⁡θ=725\cos\theta=\dfrac{7}{25}=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}-/+=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=3, find the value of sec⁡θ+tan⁡θsec⁡θ−tan⁡θ\dfrac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}+/-.

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

Then cos⁡θ=45\cos\theta=\dfrac{4}{5}=4/5, sec⁡θ=54\sec\theta=\dfrac{5}{4}=5/4, tan⁡θ=34\tan\theta=\dfrac{3}{4}=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+/-=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}R= R/ 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=√PR^2-PQ^2=√25^2-24^2=√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}R=24/25, cos⁡R=725\cos R=\dfrac{7}{25}R=7/25, tan⁡R=247\tan R=\dfrac{24}{7}R=24/7,

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

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Q12Long AnswerHOTS5 marks

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

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(i) cot⁡θ=adjacentopposite=158\cot\theta=\dfrac{\text{adjacent}}{\text{opposite}}=\dfrac{15}{8}=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=√15^2+8^2=√225+64=√289=17.

sin⁡θ=817\sin\theta=\dfrac{8}{17}=8/17, cos⁡θ=1517\cos\theta=\dfrac{15}{17}=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+)(1-)=1-^2=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+)(1-)=1-^2=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-based 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\theta be the angle the ladder makes with the ground.

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

(ii) Write sin⁡θ\sin\theta and cos⁡θ\cos\theta.

(iii) Find tan⁡θ\tan\theta.

(iv) Show that sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1^2+^2=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}=√9^2+12^2=√81+144=√225=15 m.

(ii) sin⁡θ=oppositehypotenuse=1215=45\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{12}{15}=\dfrac{4}{5}=opposite/hypotenuse=12/15=4/5; cos⁡θ=adjacenthypotenuse=915=35\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{9}{15}=\dfrac{3}{5}=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}=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}=1^2+^2=(4/5)^2+(3/5)^2=16/25+9/25=25/25=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

Frequently asked questions

  • Are these Trigonometrical Ratios important questions free?
    Yes. All 13 ICSE Class 9 Maths important questions for Trigonometrical Ratios are free, with full model answers and no login required.
  • Do these Trigonometrical Ratios questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 9 Maths, so nothing here is outside the current course.
  • How should I practise the Trigonometrical Ratios important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Trigonometrical Ratios?
    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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