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Solution of Right Triangles — ICSE Class 9 Maths Important Questions

13 ICSE Class 9 Maths practice questions on Solution of Right Triangles, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 18 October 2026

Last updated 25 October 2026

13
Questions
5
Topics
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Key concepts
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With answers
Quick answer

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,\cossin ,cos or tan⁡\tantan of a given angle as a ratio of sides, substitute the standard value, and solve. Typical answers use tan⁡45∘=1\tan45^\circ=1tan 45°=1, sin⁡30∘=12\sin30^\circ=\dfrac12sin 30°=1/2, tan⁡60∘=3\tan60^\circ=\sqrt3tan 60°=√3.

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

About Solution of Right Triangles

Within 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 valuesSimple 2-D problems on one right-angled triangle

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,\cossin ,cos or tan⁡\tantan 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}}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; 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^\circtan θ=1⇒ θ=45°, sin⁡θ=12⇒θ=30∘\sin\theta=\dfrac12\Rightarrow\theta=30^\circsin θ=1/2⇒ θ=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°-θ.

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Important questions with answers

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

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

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

  1. (a)

    5 cm5\text{ cm}5 cm

  2. (b)

    10 cm10\text{ cm}10 cm

  3. (c)

    53 cm5\sqrt3\text{ cm}5√3 cm

  4. (d)

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

Show model answer

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}sin 30°=AB/AC⇒ AB=AC sin 30°=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}}=1tan θ=opp/adj=1, so θ=45∘\theta=45^\circθ=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^\circ∠ A=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}6√3 cm

  3. (c)

    33 cm3\sqrt3\text{ cm}3√3 cm

  4. (d)

    12 cm12\text{ cm}12 cm

Show model answer

Answer: (b) 63 cm6\sqrt3\text{ cm}6√3 cm.

For ∠A=60∘\angle A=60^\circ∠ A=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}tan 60°=BC/AB⇒ BC=6 tan 60°=6√3 cm.

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Q4MCQHOTS1 mark

In a right triangle right-angled at BBB, tan⁡A=3\tan A=\sqrt3tan A=√3. The measure of ∠C\angle C∠ C 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: (a) 30∘30^\circ30°.

tan⁡A=3⇒∠A=60∘\tan A=\sqrt3\Rightarrow\angle A=60^\circtan A=√3⇒ ∠ A=60°. Since ∠A+∠C=90∘\angle A+\angle C=90^\circ∠ A+∠ C=90°, ∠C=90∘−60∘=30∘\angle C=90^\circ-60^\circ=30^\circ∠ C=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^\circ∠ C=45° and BC=7 cmBC=7\text{ cm}BC=7 cm. Find ABABAB.

Show model answer

For ∠C=45∘\angle C=45^\circ∠ C=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}tan 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 P∠ P and the hypotenuse PRPRPR.

Show model answer

tan⁡P=QRPQ=55=1⇒∠P=45∘\tan P=\dfrac{QR}{PQ}=\dfrac{5}{5}=1\Rightarrow\angle P=45^\circtan P=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=√(PQ²+QR²)=√(5²+5²)=√50=5√2 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^\circ∠ A=30° and BC=4 cmBC=4\text{ cm}BC=4 cm. Find ABABAB and the hypotenuse ACACAC.

Show model answer

For ∠A=30∘\angle A=30^\circ∠ A=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}tan 30°=BC/AB⇒ 1/√3=4/AB⇒ AB=4√3 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}sin 30°=BC/AC⇒ 1/2=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=6√3 cm. Find ∠A\angle A∠ A, ∠C\angle C∠ C 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^\circtan A=BC/AB=6√3/6=√3⇒ ∠ A=60°.

∠C=90∘−∠A=90∘−60∘=30∘\angle C=90^\circ-\angle A=90^\circ-60^\circ=30^\circ∠ C=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=√(AB²+BC²)=√(6²+(6√3)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^\circ∠ C=60°. Find ABABAB and BCBCBC, taking 3=1.73\sqrt3=1.73√3=1.73.

Show model answer

For ∠C=60∘\angle C=60^\circ∠ C=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}sin 60°=AB/AC⇒ √3/2=AB/14⇒ AB=14× √3/2=7√3=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}cos 60°=BC/AC⇒ 1/2=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^\circ∠ A=45° and AB=10 cmAB=10\text{ cm}AB=10 cm. Solve the triangle completely (find ∠C\angle C∠ C, 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^\circ∠ A=45°.

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

Side BCBCBC: For ∠A=45∘\angle A=45^\circ∠ A=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}tan 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=√(AB²+BC²)=√(10²+10²)=√200=10√2 cm≈14.14 cm.

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Q12Long AnswerModerate5 marks

A ladder 101010 m long rests against a vertical wall and makes an angle of 60∘60^\circ60° with the level ground. Find (i) the height of the wall reached by the ladder, (ii) the distance of the foot of the ladder from the wall, (iii) the angle the ladder makes with the wall. Take 3=1.73\sqrt3=1.73√3=1.73.

Show model answer

The ladder, wall and ground form a right triangle with the ladder (101010 m) as hypotenuse.

(i) sin⁡60∘=h10⇒h=10×32=53=5×1.73=8.65\sin60^\circ=\dfrac{h}{10}\Rightarrow h=10\times\dfrac{\sqrt3}{2}=5\sqrt3=5\times1.73=8.65sin 60°=h/10⇒ h=10× √3/2=5√3=5×1.73=8.65 m.

(ii) cos⁡60∘=d10⇒d=10×12=5\cos60^\circ=\dfrac{d}{10}\Rightarrow d=10\times\dfrac12=5cos 60°=d/10⇒ d=10× 1/2=5 m.

(iii) The two acute angles add up to 90∘90^\circ90°, so the angle with the wall is 90∘−60∘=30∘90^\circ-60^\circ=30^\circ90°-60°=30°.

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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=1.73√3=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 A∠ A (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^\circ∠ C=30°. Here ABABAB is opposite ∠C\angle C∠ C and BCBCBC is adjacent.

(i) The tangent ratio links them: tan⁡∠C=ABBC\tan\angle C=\dfrac{AB}{BC}tan ∠ 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}tan 30°=AB/BC⇒ 1/√3=AB/12⇒ AB=12/√3=12/1.73≈6.93 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}cos 30°=BC/CA⇒ √3/2=12/CA⇒ CA=24/√3=24/1.73≈13.87 m.

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

  • What does it mean to solve a right triangle?
    To find all its unknown sides and acute angles from the given ones, using trigonometric ratios of standard angles and the fact that the two acute angles add up to 90∘90^\circ90°. This page on Solution of Right Triangles has 13 of them, each with a worked answer.
  • How do you find the sides of a right triangle with a 30° angle?
    Pick the ratio that links what you know to what you want. With ∠A=30∘\angle A=30^\circ∠ A=30° and BC=4BC=4BC=4 cm opposite it: AC=4sin⁡30∘=8AC=\tfrac{4}{\sin30^\circ}=8AC=4/(sin 30°)=8 cm and AB=4tan⁡30∘=43AB=\tfrac{4}{\tan30^\circ}=4\sqrt3AB=4/(tan 30°)=4√3 cm.
  • How do you find an angle when two sides are known?
    Form the ratio and match it to a standard value. If AB=6AB=6AB=6 and BC=63BC=6\sqrt3BC=6√3 with the right angle at BBB, then tan⁡A=3\tan A=\sqrt3tan A=√3, so ∠A=60∘\angle A=60^\circ∠ A=60°, ∠C=30∘\angle C=30^\circ∠ C=30° and AC=12AC=12AC=12.
  • How high up a wall does a 10 m ladder at 60° reach?
    Height =10sin⁡60∘=53≈8.66=10\sin60^\circ=5\sqrt3\approx8.66=10 sin 60°=5√3≈8.66 m, and the foot is 10cos⁡60∘=510\cos60^\circ=510 cos 60°=5 m from the wall. Both come from the ladder being the hypotenuse. Draw the triangle with the angle at the ground before choosing a ratio.
  • Does Solution of Right Triangles lead to Heights and Distances?
    Yes. ICSE Class 10's Heights and Distances (Chapter 22) applies the same methods with angles of elevation and depression. In Class 9 this is Chapter 24, followed by Complementary Angles (Chapter 25).
  • Which ratio should I choose when solving a right triangle?
    Use tan⁡\tantan when the known and unknown sides are the two legs, and sin⁡\sinsin or cos⁡\coscos when the hypotenuse is involved. Choosing a ratio that puts the unknown in the numerator avoids dividing by surds.
  • Can I download Solution of Right Triangles ICSE questions?
    Yes. Enter your email in the free PDF box on this page to open all 13 right-triangle questions, the pole case study included, as a printable worksheet with answers.

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