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Solution of Right Triangles — ICSE Class 9 Maths Important Questions

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

13
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6
Question types
32
Total marks
₹0
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Quick answer

High-yield ICSE Class 9 questions on the Solution of Right Triangles use trigonometrical ratios of standard angles to find unknown sides or angles of a right triangle. Set up sin⁡,cos⁡\sin,\cos, or tan⁡\tan of a given angle as a ratio of sides, substitute the standard value, and solve. Typical answers use tan⁡45∘=1\tan45^\circ=145^=1, sin⁡30∘=12\sin30^\circ=\dfrac1230^=12, tan⁡60∘=3\tan60^\circ=\sqrt360^=3.

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

About Solution of Right Triangles

In the ICSE Class 9 Maths chapter Solution of Right Triangles you find the unknown sides and angles of a right-angled triangle when enough information is given, using the trigonometrical ratios of standard angles. Selina-aligned questions provide one side and one acute angle, or two sides, and ask you to compute the remaining parts. You choose the ratio that links the known and unknown quantities and substitute exact standard-angle values.

Choosing the correct ratio (sin, cos, tan)Finding a side given a side and an angleFinding an angle given two sidesUsing standard-angle values in solving trianglesTwo-triangle and multi-step problems

Key concepts & formulas

Solving a right triangle

To 'solve' a right triangle is to find all its unknown sides and acute angles. Choose sin⁡,cos⁡\sin,\cos, or tan⁡\tan so that it relates the known side/angle to the unknown you want.

Ratio selection

Use 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; pick the one containing the known and required sides.

Finding an angle

Given two sides, form the ratio and match it to a standard value, e.g. tan⁡θ=1⇒θ=45∘\tan\theta=1\Rightarrow\theta=45^\circ=1=45^, sin⁡θ=12⇒θ=30∘\sin\theta=\dfrac12\Rightarrow\theta=30^\circ=12=30^.

Angle sum

In a right triangle the two acute angles add to 90∘90^\circ90^, so once one acute angle is known the other is 90∘−θ90^\circ-\theta90^-.

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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

In a right triangle right-angled at BBB, if ∠C=30∘\angle C=30^\circC=30^ and the hypotenuse AC=10 cmAC=10\text{ cm}AC=10 cm, then the side ABABAB opposite ∠C\angle CC is:

  1. (a)

    5 cm5\text{ cm}5 cm

  2. (b)

    10 cm10\text{ cm}10 cm

  3. (c)

    53 cm5\sqrt3\text{ cm}53 cm

  4. (d)

    103 cm\dfrac{10}{\sqrt3}\text{ cm}10/3 cm

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Answer: (a) 5 cm5\text{ cm}5 cm.

sin⁡30∘=ABAC⇒AB=ACsin⁡30∘=10×12=5 cm\sin30^\circ=\dfrac{AB}{AC}\Rightarrow AB=AC\sin30^\circ=10\times\dfrac{1}{2}=5\text{ cm}30^=AB/AC AB=AC30^=10×1/2=5 cm.

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Q2MCQEasy1 mark

In a right triangle, if the two legs are equal, each acute angle is:

  1. (a)

    30∘30^\circ30^

  2. (b)

    45∘45^\circ45^

  3. (c)

    60∘60^\circ60^

  4. (d)

    90∘90^\circ90^

Show model answer

Answer: (b) 45∘45^\circ45^.

Equal legs give tan⁡θ=oppadj=1\tan\theta=\dfrac{\text{opp}}{\text{adj}}=1=opp/adj=1, so θ=45∘\theta=45^\circ=45^ (and both acute angles are 45∘45^\circ45^).

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Q3MCQModerate1 mark

In right triangle ABCABCABC (right-angled at BBB), ∠A=60∘\angle A=60^\circA=60^ and AB=6 cmAB=6\text{ cm}AB=6 cm. The length of BCBCBC is:

  1. (a)

    6 cm6\text{ cm}6 cm

  2. (b)

    63 cm6\sqrt3\text{ cm}63 cm

  3. (c)

    33 cm3\sqrt3\text{ cm}33 cm

  4. (d)

    12 cm12\text{ cm}12 cm

Show model answer

Answer: (b) 63 cm6\sqrt3\text{ cm}63 cm.

For ∠A=60∘\angle A=60^\circA=60^, BCBCBC is opposite and ABABAB is adjacent, so tan⁡60∘=BCAB⇒BC=6tan⁡60∘=63 cm\tan60^\circ=\dfrac{BC}{AB}\Rightarrow BC=6\tan60^\circ=6\sqrt3\text{ cm}60^=BC/AB BC=660^=63 cm.

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Q4MCQHOTS1 mark

In a right triangle right-angled at BBB, tan⁡A=3\tan A=\sqrt3A=3. The measure of ∠C\angle CC is:

  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: (a) 30∘30^\circ30^.

tan⁡A=3⇒∠A=60∘\tan A=\sqrt3\Rightarrow\angle A=60^\circA=3 A=60^. Since ∠A+∠C=90∘\angle A+\angle C=90^\circA+ C=90^, ∠C=90∘−60∘=30∘\angle C=90^\circ-60^\circ=30^\circC=90^-60^=30^.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): In a right triangle, if one acute angle is 50∘50^\circ50^ then the other acute angle is 40∘40^\circ40^.

Reason (R): The sum of the three angles of a triangle is 180∘180^\circ180^, and one angle of a right triangle is 90∘90^\circ90^.

  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

Show model answer

Answer: (a) The two acute angles sum to 180∘−90∘=90∘180^\circ-90^\circ=90^\circ180^-90^=90^, so the other is 90∘−50∘=40∘90^\circ-50^\circ=40^\circ90^-50^=40^. 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, ∠C=45∘\angle C=45^\circC=45^ and BC=7 cmBC=7\text{ cm}BC=7 cm. Find ABABAB.

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For ∠C=45∘\angle C=45^\circC=45^, ABABAB is opposite and BCBCBC is adjacent.

tan⁡45∘=ABBC⇒1=AB7⇒AB=7 cm\tan45^\circ=\dfrac{AB}{BC}\Rightarrow 1=\dfrac{AB}{7}\Rightarrow AB=7\text{ cm}45^=AB/BC 1=AB/7 AB=7 cm.

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Q7Very ShortModerate2 marks

In right triangle PQRPQRPQR, right-angled at QQQ, PQ=QR=5 cmPQ=QR=5\text{ cm}PQ=QR=5 cm. Find ∠P\angle PP and the hypotenuse PRPRPR.

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tan⁡P=QRPQ=55=1⇒∠P=45∘\tan P=\dfrac{QR}{PQ}=\dfrac{5}{5}=1\Rightarrow\angle P=45^\circP=QR/PQ=5/5=1 P=45^.

Hypotenuse PR=PQ2+QR2=52+52=50=52 cmPR=\sqrt{PQ^2+QR^2}=\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt2\text{ cm}PR=√PQ^2+QR^2=√5^2+5^2=√50=52 cm.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

In right triangle ABCABCABC, right-angled at BBB, ∠A=30∘\angle A=30^\circA=30^ and BC=4 cmBC=4\text{ cm}BC=4 cm. Find ABABAB and the hypotenuse ACACAC.

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For ∠A=30∘\angle A=30^\circA=30^: BCBCBC (opposite) =4 cm=4\text{ cm}=4 cm.

tan⁡30∘=BCAB⇒13=4AB⇒AB=43 cm\tan30^\circ=\dfrac{BC}{AB}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{4}{AB}\Rightarrow AB=4\sqrt3\text{ cm}30^=BC/AB1/3=4/AB AB=43 cm.

sin⁡30∘=BCAC⇒12=4AC⇒AC=8 cm\sin30^\circ=\dfrac{BC}{AC}\Rightarrow\dfrac{1}{2}=\dfrac{4}{AC}\Rightarrow AC=8\text{ cm}30^=BC/AC1/2=4/AC AC=8 cm.

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Q9Short AnswerModerate3 marks

In right triangle ABCABCABC, right-angled at BBB, AB=6 cmAB=6\text{ cm}AB=6 cm and BC=63 cmBC=6\sqrt3\text{ cm}BC=63 cm. Find ∠A\angle AA, ∠C\angle CC and the hypotenuse ACACAC.

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tan⁡A=BCAB=636=3⇒∠A=60∘\tan A=\dfrac{BC}{AB}=\dfrac{6\sqrt3}{6}=\sqrt3\Rightarrow\angle A=60^\circA=BC/AB=63/6=3 A=60^.

∠C=90∘−∠A=90∘−60∘=30∘\angle C=90^\circ-\angle A=90^\circ-60^\circ=30^\circC=90^- A=90^-60^=30^.

AC=AB2+BC2=62+(63)2=36+108=144=12 cmAC=\sqrt{AB^2+BC^2}=\sqrt{6^2+(6\sqrt3)^2}=\sqrt{36+108}=\sqrt{144}=12\text{ cm}AC=√AB^2+BC^2=√6^2+(63)^2=√36+108=√144=12 cm.

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Q10Short AnswerHOTS3 marks

A right triangle ABCABCABC is right-angled at BBB. Its hypotenuse AC=14 cmAC=14\text{ cm}AC=14 cm and ∠C=60∘\angle C=60^\circC=60^. Find ABABAB and BCBCBC, taking 3=1.73\sqrt3=1.733=1.73.

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For ∠C=60∘\angle C=60^\circC=60^: ABABAB is opposite, BCBCBC is adjacent, ACACAC is hypotenuse.

sin⁡60∘=ABAC⇒32=AB14⇒AB=14×32=73=7×1.73=12.11 cm\sin60^\circ=\dfrac{AB}{AC}\Rightarrow\dfrac{\sqrt3}{2}=\dfrac{AB}{14}\Rightarrow AB=14\times\dfrac{\sqrt3}{2}=7\sqrt3=7\times1.73=12.11\text{ cm}60^=AB/AC3/2=AB/14 AB=14×3/2=73=7×1.73=12.11 cm.

cos⁡60∘=BCAC⇒12=BC14⇒BC=7 cm\cos60^\circ=\dfrac{BC}{AC}\Rightarrow\dfrac{1}{2}=\dfrac{BC}{14}\Rightarrow BC=7\text{ cm}60^=BC/AC1/2=BC/14 BC=7 cm.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

In the figure, ABCABCABC is right-angled at BBB with ∠A=45∘\angle A=45^\circA=45^ and AB=10 cmAB=10\text{ cm}AB=10 cm. Solve the triangle completely (find ∠C\angle CC, BCBCBC and ACACAC).

ICSE Class 9 Maths — Solution of Right Triangles: In the figure, ABC is right-angled at B with \angle A=45^\circ and AB=10\text{ cm}. Solve the triangle completely (find \angle C,
Show model answer

The right angle is at BBB, and ∠A=45∘\angle A=45^\circA=45^.

Angle CCC: ∠C=90∘−∠A=90∘−45∘=45∘\angle C=90^\circ-\angle A=90^\circ-45^\circ=45^\circC=90^- A=90^-45^=45^.

Side BCBCBC: For ∠A=45∘\angle A=45^\circA=45^, BCBCBC is opposite and ABABAB is adjacent.

tan⁡45∘=BCAB⇒1=BC10⇒BC=10 cm\tan45^\circ=\dfrac{BC}{AB}\Rightarrow 1=\dfrac{BC}{10}\Rightarrow BC=10\text{ cm}45^=BC/AB 1=BC/10 BC=10 cm.

Hypotenuse ACACAC: AC=AB2+BC2=102+102=200=102 cm≈14.14 cmAC=\sqrt{AB^2+BC^2}=\sqrt{10^2+10^2}=\sqrt{200}=10\sqrt2\text{ cm}\approx14.14\text{ cm}AC=√AB^2+BC^2=√10^2+10^2=√200=102 cm14.14 cm.

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Q12Long AnswerHOTS5 marks

In the figure, BDBDBD is perpendicular to ACACAC. ∠A=30∘\angle A=30^\circA=30^, ∠C=45∘\angle C=45^\circC=45^ and BD=6 cmBD=6\text{ cm}BD=6 cm. Find (i) ADADAD, (ii) DCDCDC and (iii) the length ACACAC. Take 3=1.73\sqrt3=1.733=1.73.

ICSE Class 9 Maths — Solution of Right Triangles: In the figure, BD is perpendicular to AC. \angle A=30^\circ, \angle C=45^\circ and BD=6\text{ cm}. Find (i) AD, (ii) DC and (iii)
Show model answer

BD⊥ACBD\perp ACBD AC, so triangles ABDABDABD and CBDCBDCBD are both right-angled at DDD, with BD=6 cmBD=6\text{ cm}BD=6 cm.

(i) In right triangle ABDABDABD, ∠A=30∘\angle A=30^\circA=30^, BDBDBD opposite, ADADAD adjacent.

tan⁡30∘=BDAD⇒13=6AD⇒AD=63=6×1.73=10.38 cm\tan30^\circ=\dfrac{BD}{AD}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{6}{AD}\Rightarrow AD=6\sqrt3=6\times1.73=10.38\text{ cm}30^=BD/AD1/3=6/AD AD=63=6×1.73=10.38 cm.

(ii) In right triangle CBDCBDCBD, ∠C=45∘\angle C=45^\circC=45^, BDBDBD opposite, DCDCDC adjacent.

tan⁡45∘=BDDC⇒1=6DC⇒DC=6 cm\tan45^\circ=\dfrac{BD}{DC}\Rightarrow 1=\dfrac{6}{DC}\Rightarrow DC=6\text{ cm}45^=BD/DC 1=6/DC DC=6 cm.

(iii) AC=AD+DC=10.38+6=16.38 cmAC=AD+DC=10.38+6=16.38\text{ cm}AC=AD+DC=10.38+6=16.38 cm.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A vertical pole ABABAB stands on level ground. From a point CCC on the ground, the foot BBB is 12 m12\text{ m}12 m away, and the pole makes a right angle with the ground at BBB. The line CACACA from CCC to the top AAA makes an angle of 30∘30^\circ30^ with the ground. Take 3=1.73\sqrt3=1.733=1.73.

(i) Which trigonometric ratio links the height ABABAB, the base BCBCBC and the angle 30∘30^\circ30^?

(ii) Find the height ABABAB of the pole.

(iii) Find the length CACACA.

(iv) State the size of ∠A\angle AA (the angle at the top of the pole in triangle ABCABCABC).

Show model answer

Triangle ABCABCABC is right-angled at BBB, with BC=12 mBC=12\text{ m}BC=12 m and ∠C=30∘\angle C=30^\circC=30^. Here ABABAB is opposite ∠C\angle CC and BCBCBC is adjacent.

(i) The tangent ratio links them: tan⁡∠C=ABBC\tan\angle C=\dfrac{AB}{BC}C=AB/BC.

(ii) tan⁡30∘=ABBC⇒13=AB12⇒AB=123=121.73≈6.93 m\tan30^\circ=\dfrac{AB}{BC}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{AB}{12}\Rightarrow AB=\dfrac{12}{\sqrt3}=\dfrac{12}{1.73}\approx6.93\text{ m}30^=AB/BC1/3=AB/12 AB=12/3=12/1.736.93 m.

(iii) cos⁡30∘=BCCA⇒32=12CA⇒CA=243=241.73≈13.87 m\cos30^\circ=\dfrac{BC}{CA}\Rightarrow\dfrac{\sqrt3}{2}=\dfrac{12}{CA}\Rightarrow CA=\dfrac{24}{\sqrt3}=\dfrac{24}{1.73}\approx13.87\text{ m}30^=BC/CA3/2=12/CA CA=24/3=24/1.7313.87 m.

(iv) ∠A=90∘−∠C=90∘−30∘=60∘\angle A=90^\circ-\angle C=90^\circ-30^\circ=60^\circA=90^- C=90^-30^=60^.

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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 Solution of Right Triangles important questions free?
    Yes. All 13 ICSE Class 9 Maths important questions for Solution of Right Triangles are free, with full model answers and no login required.
  • Do these Solution of Right Triangles 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 Solution of Right Triangles 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 Solution of Right Triangles?
    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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