NDA Maths · Properties of Triangle

Sine & Cosine Rules — Solving Triangles

The sine rule links each side to the sine of its opposite angle (and to the circumradius); the cosine rule links one side to the other two and their included angle. Together they let you find any missing side, angle, or area.

Why this matters

The chapter's largest and hardest pocket (29 PYQs, 13 HARD). Once you fix the notation — side a opposite angle A — almost every question is 'which rule fits the given data?': sine rule when you have an angle and its opposite side, cosine rule when you have all three sides or two sides and the included angle.

Concept 1 of 7

Triangle Notation & Basic Relations

Intuition

Everything in this chapter rides on one convention: name each side with the lower-case letter of the angle OPPOSITE it. With that fixed, the angle sum and a few derived quantities (semi-perimeter, circumradius, inradius) give you a common language for every formula.

Definition

For triangle ABCABC, the side opposite vertex AA is a=BCa = BC, opposite BB is b=CAb = CA, opposite CC is c=ABc = AB. Standard quantities:

  • Angle sum: A+B+C=πA + B + C = \pi (so any one angle is determined by the other two).
  • Semi-perimeter: s=a+b+c2s = \dfrac{a+b+c}{2}.
  • **Circumradius RR (radius of the circle through all three vertices) and inradius rr** (radius of the circle touching all three sides).
  • Largest side faces the largest angle, and the longest side is opposite the obtuse angle if there is one — a quick orientation check.

Angle sum & semi-perimeter

A+B+C=π,s=a+b+c2A + B + C = \pi, \qquad s = \dfrac{a+b+c}{2}
ABCabca opposite A · b opposite B · c opposite C

Worked example

In ABC\triangle ABC, A=50A = 50^\circ and B=60B = 60^\circ. Which side is the longest?
  1. C=1805060=70C = 180^\circ - 50^\circ - 60^\circ = 70^\circ is the largest angle.
  2. The longest side faces the largest angle, so side cc (opposite CC) is longest.
Answer:Side cc (=AB= AB).
Practice this concept2 quick reps

Practice — Level 1 (2 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    If A=90A = 90^\circ, which side is the hypotenuse?
  2. 2.
    Sides 5,7,85, 7, 8: which angle is largest?

Concept 2 of 7

The Sine Rule

Intuition

The ratio of a side to the sine of its opposite angle is the same for all three sides — and equals the diameter of the circumcircle. Use it whenever you know an angle and the side facing it.

Definition

For any triangle,

asinA=bsinB=csinC=2R.\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R.
Consequences: a=2RsinAa = 2R\sin A, so the sides are proportional to the sines of the opposite angles; the perimeter is 2R(sinA+sinB+sinC)2R(\sin A + \sin B + \sin C). Use the sine rule when you have an angle and its opposite side (plus one more angle or side). Beware the ambiguous case: knowing two sides and a non-included angle can give two valid triangles.

Sine rule

asinA=bsinB=csinC=2R\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R

Worked example

In ABC\triangle ABC, A=30A = 30^\circ, a=5a = 5. Find the circumradius RR.
  1. Sine rule: asinA=2R\dfrac{a}{\sin A} = 2R.
  2. 2R=5sin30=51/2=102R = \dfrac{5}{\sin 30^\circ} = \dfrac{5}{1/2} = 10.
Answer:R=5R = 5.

From the bank · past-year question

Example 2Properties of TriangleEASY
In triangle ABC if a=2a=2, b=3b=3 and sinA=2/3\sin A = 2/3, then what is angle B?

[Q39 · Apr · 2018]

Side over sine of its OWN opposite angle

The sine rule pairs each side with the angle facing it: asinA\frac{a}{\sin A}, never asinB\frac{a}{\sin B}. Pairing a side with the wrong angle is the classic slip.

Concept 3 of 7

The Cosine Rule

Intuition

The cosine rule is the Pythagorean theorem with a correction term for the angle. Use it when you have all three sides (to find an angle) or two sides and the angle between them (to find the third side).

Definition

For any triangle,

c2=a2+b22abcosC,cosC=a2+b2c22ab,c^2 = a^2 + b^2 - 2ab\cos C, \qquad \cos C = \dfrac{a^2 + b^2 - c^2}{2ab},
and cyclically for A,BA, B. The sign of the cosine reveals the angle: cosC>0\cos C > 0 acute, =0= 0 right (c2=a2+b2c^2 = a^2 + b^2), <0< 0 obtuse. Once an angle's cosine is known, double/triple-angle formulas give cos2C\cos 2C, cos3C\cos 3C, etc.

Cosine rule

cosC=a2+b2c22ab\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}

Worked example

In ABC\triangle ABC with a=7, b=8, c=9a = 7,\ b = 8,\ c = 9, find cosA\cos A.
  1. cosA=b2+c2a22bc=64+8149289\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{64 + 81 - 49}{2\cdot 8 \cdot 9}.
  2. =96144= \dfrac{96}{144}.
Answer:cosA=23\cos A = \dfrac{2}{3}.

From the bank · past-year question

Example 3Properties of TriangleHARD
In a triangle ABC, a=4,b=3,c=2a = 4, b = 3, c = 2. What is cos3C\cos 3C equal to?

[Q50 · Sep · 2022]

Put the opposite side as the one being squared on the left

For angle CC, the formula has c2c^2 (the side opposite CC) isolated and 2abcosC-2ab\cos C. Mixing up which side is opposite the angle flips the sign of the cosine and the verdict on acute/obtuse.

Concept 4 of 7

Determining the Nature of a Triangle

Intuition

Many questions hand you a relation among the sides or angles and ask 'what kind of triangle is it?'. The cosine rule (sign of a cosine) settles right vs obtuse; symmetric side conditions force equilateral or isosceles.

Definition

Tests for the type of triangle:

  • Right-angled: a2+b2=c2a^2 + b^2 = c^2 for the largest side cc; equivalently a cosine is 0, or cos2A+cos2B+cos2C=1\cos^2 A + \cos^2 B + \cos^2 C = 1, or sin2A+sin2B+sin2C=2\sin^2 A + \sin^2 B + \sin^2 C = 2.
  • Obtuse: the largest side satisfies c2>a2+b2c^2 > a^2 + b^2 (its cosine is negative).
  • Equilateral: symmetric conditions like acosA=bcosB=ccosCa\cos A = b\cos B = c\cos C force a=b=ca = b = c.
  • A condition like c2=a2+b2+abc^2 = a^2 + b^2 + ab gives cosC=12\cos C = -\tfrac12, so C=120C = 120^\circ.

Right-angle tests

c2=a2+b2    C=90    cos2A+cos2B+cos2C=1c^2 = a^2 + b^2 \iff C = 90^\circ \iff \cos^2 A + \cos^2 B + \cos^2 C = 1

Worked example

The sides of a triangle are p, p, p2p,\ p,\ p\sqrt{2}. What is its nature?
  1. Check the largest side p2p\sqrt2: (p2)2=2p2(p\sqrt2)^2 = 2p^2 and p2+p2=2p2p^2 + p^2 = 2p^2.
  2. Since (p2)2=p2+p2(p\sqrt2)^2 = p^2 + p^2, the angle opposite it is 9090^\circ; the two equal sides make it isosceles.
Answer:Right-angled isosceles triangle.

From the bank · past-year question

Example 4Properties of TriangleMODERATE
In triangle ABCABC, if sin2A+sin2B+sin2Ccos2A+cos2B+cos2C=2\dfrac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2, then the triangle is

[Q39 · Sep · 2017]

Concept 5 of 7

Area of a Triangle

Intuition

Pick the area formula that matches your data: two sides and the included angle, or all three sides (Heron), or the circumradius, or the inradius and semi-perimeter. They all give the same Δ.

Definition

Let Δ\Delta be the area:

  • Two sides + included angle: Δ=12absinC\Delta = \tfrac12 ab\sin C.
  • Three sides (Heron): Δ=s(sa)(sb)(sc)\Delta = \sqrt{s(s-a)(s-b)(s-c)}.
  • Circumradius: Δ=abc4R\Delta = \dfrac{abc}{4R}.
  • Inradius: Δ=rs\Delta = r s.

Area formulas

Δ=12absinC=s(sa)(sb)(sc)=abc4R=rs\Delta = \tfrac12 ab\sin C = \sqrt{s(s-a)(s-b)(s-c)} = \dfrac{abc}{4R} = rs

Worked example

Find the area of a triangle with sides 13, 14, 1513,\ 14,\ 15.
  1. s=13+14+152=21s = \tfrac{13+14+15}{2} = 21.
  2. Heron: Δ=21(2113)(2114)(2115)=21876=7056\Delta = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21\cdot 8\cdot 7\cdot 6} = \sqrt{7056}.
Answer:Δ=84\Delta = 84.

From the bank · past-year question

Example 5Properties of TriangleEASY
What is the area of the triangle ABCABC with sides a=10a = 10 cm, c=4c = 4 cm and angle B=30°B = 30°?

[Q34 · Apr · 2021]

Concept 6 of 7

Angle Ratios ↔ Side Ratios

Intuition

Because sides are proportional to the sines of opposite angles, any condition on the ANGLES (in AP, in a given ratio) converts to a condition on the SIDES, and vice versa. Use the angle sum to pin the angles, then the sine rule to get the sides.

Definition

  • Angles in AP: 2B=A+C2B = A + C together with A+B+C=πA+B+C = \pi forces the middle angle B=60B = 60^\circ.
  • Angles in a given ratio (e.g. 1:2:31:2:3): split 180180^\circ accordingly, then sides sinA:sinB:sinC\propto \sin A : \sin B : \sin C.
  • Given a side ratio, the sine rule recovers the angles: bc=sinBsinC\dfrac{b}{c} = \dfrac{\sin B}{\sin C}.

Sides proportional to sines

a:b:c=sinA:sinB:sinCa : b : c = \sin A : \sin B : \sin C

Worked example

The angles of a triangle are in the ratio 1:2:31 : 2 : 3. Find the ratio of the sides.
  1. Angles: 30,60,9030^\circ, 60^\circ, 90^\circ (they sum to 180180^\circ).
  2. Sides sin30:sin60:sin90=12:32:1\propto \sin 30^\circ : \sin 60^\circ : \sin 90^\circ = \tfrac12 : \tfrac{\sqrt3}{2} : 1.
Answer:1:3:21 : \sqrt{3} : 2.

From the bank · past-year question

Example 6Properties of TriangleMODERATE
If the angles of a triangle ABCABC are in AP and b:c=3:2b : c = \sqrt{3} : \sqrt{2}, then what is the measure of angle AA ?

[Q62 · Sep · 2019]

Concept 7 of 7

Sine Rule in Geometric Configurations

Intuition

When a figure is split into sub-triangles (a cevian, a quadrilateral diagonal, an angle divided into two parts), apply the sine rule inside the relevant sub-triangle. The shared side or angle ties the pieces together.

Definition

In a sub-triangle, the sine rule still holds. Useful moves:

  • In ABD\triangle ABD, ABsin(ADB)=ADsin(ABD)=BDsin(DAB)\dfrac{AB}{\sin(\angle ADB)} = \dfrac{AD}{\sin(\angle ABD)} = \dfrac{BD}{\sin(\angle DAB)}, and DAB=π(the other two)\angle DAB = \pi - (\text{the other two}).
  • An angle α\alpha split into parts A,BA, B with a tangent ratio: use tanAtanBtanA+tanB=sin(AB)sin(A+B)\dfrac{\tan A - \tan B}{\tan A + \tan B} = \dfrac{\sin(A-B)}{\sin(A+B)} (componendo–dividendo on tanA:tanB\tan A : \tan B).

Sine rule in a sub-triangle

ABsin(ADB)=ADsin(ABD)\dfrac{AB}{\sin(\angle ADB)} = \dfrac{AD}{\sin(\angle ABD)}

Worked example

In ABD\triangle ABD, ADB=θ\angle ADB = \theta and ABD=α\angle ABD = \alpha. Show ADsinθ=ABsinαAD\sin\theta = AB\sin\alpha.
  1. Sine rule in ABD\triangle ABD: ABsinθ=ADsinα\dfrac{AB}{\sin\theta} = \dfrac{AD}{\sin\alpha}.
  2. Cross-multiplying gives ABsinα=ADsinθAB\sin\alpha = AD\sin\theta.
Answer:ADsinθ=ABsinαAD\sin\theta = AB\sin\alpha.

From the bank · past-year question

Example 7Properties of TriangleHARD
Directions for the following two (02) items: Read the following information and answer the two items that follow: ABCD is a trapezium such that AB and CD are parallel and BC is perpendicular to them. Let ∠ADB = θ, ∠ABD = α, BC = p and CD = q.
Consider the following: 1. ADsinθ=ABsinαAD\sin\theta = AB\sin\alpha 2. BDsinθ=ABsin(θ+α)BD\sin\theta = AB\sin(\theta+\alpha). Which of the above is/are correct?

[Q21 · Apr · 2020]

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

  • Triangle Notation & Basic Relations

    Angle sum & semi-perimeter

    A+B+C=π,s=a+b+c2A + B + C = \pi, \qquad s = \dfrac{a+b+c}{2}
  • The Sine Rule

    Sine rule

    asinA=bsinB=csinC=2R\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R
  • The Cosine Rule

    Cosine rule

    cosC=a2+b2c22ab\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}
  • Determining the Nature of a Triangle

    Right-angle tests

    c2=a2+b2    C=90    cos2A+cos2B+cos2C=1c^2 = a^2 + b^2 \iff C = 90^\circ \iff \cos^2 A + \cos^2 B + \cos^2 C = 1
  • Area of a Triangle

    Area formulas

    Δ=12absinC=s(sa)(sb)(sc)=abc4R=rs\Delta = \tfrac12 ab\sin C = \sqrt{s(s-a)(s-b)(s-c)} = \dfrac{abc}{4R} = rs
  • Angle Ratios ↔ Side Ratios

    Sides proportional to sines

    a:b:c=sinA:sinB:sinCa : b : c = \sin A : \sin B : \sin C
  • Sine Rule in Geometric Configurations

    Sine rule in a sub-triangle

    ABsin(ADB)=ADsin(ABD)\dfrac{AB}{\sin(\angle ADB)} = \dfrac{AD}{\sin(\angle ABD)}

Watch out for (2)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Properties of TriangleHARD
Consider the following for the items that follow: The perimeter of a triangle ABCABC is 6 times the AM of sine of angles of the triangle. Further BC=3BC=\sqrt{3} and CA=1CA=1.
What is the perimeter of the triangle?

[Q35 · Apr · 2023]

Example 2Properties of TriangleMODERATE
The sides of a triangle ABC are AB = 3 cm, BC = 5 cm and CA = 7 cm.
Consider the following statements: (I). The triangle is obtuse-angled triangle. (II). The sum of acute angles of the triangle is also acute. Which of the statements given above is/are correct?

[Q36 · Apr · 2025]

Example 3Properties of TriangleMODERATE
In PQR\triangle PQR, R=π2\angle R = \dfrac{\pi}{2}. If tan ⁣(P2)\tan\!\left(\dfrac{P}{2}\right) and tan ⁣(Q2)\tan\!\left(\dfrac{Q}{2}\right) are the roots of the equation ax2+bx+c=0ax^2 + bx + c = 0, then which one of the following is correct?

[Q29 · Sep · 2017]

Example 4Properties of TriangleMODERATE
In a triangle ABCABC, acosA=bcosB=ccosCa\cos A=b\cos B=c\cos C. What is the area of the triangle if a=6a=6 cm?

[Q23 · Sep · 2024]

Example 5Properties of TriangleHARD
The angles AA, BB and CC of triangle ABCABC are in the ratio 3:5:43:5:4.
What is a+b+2ca+b+2c equal to?

[Q79 · Sep · 2023]

Drill every past-year question on this subtopic

29 questions from the bank — paginated, with cart and Word-export support.