MHT-CET Maths · Trigonometric Functions

Solution of a Triangle — the Sine, Cosine and Projection Rules

In a triangle with sides a, b, c opposite angles A, B, C, the sine rule links each side to its opposite angle, the cosine rule links three sides to one angle, and the projection rule writes each side as the sum of the other two sides' projections on it.

All Trigonometric Functions notes MHT-CET Maths strategy

Why this matters

45 PYQs, the larger of the two triangle pages. Sixteen are the sine rule (angle ratios to side ratios, circumradius, which triangles exist), twenty-four the cosine rule (an angle from three sides, or an angle from a relation among the sides), and five the projection rule. The rule to use is decided by what the stem gives you, so recognising the given data is most of the question.

Concept 1 of 4: The Sine Rule and the Circumradius

Every triangle sits in a circle, and each side is a chord of it seen from the opposite vertex. A chord's length is

2R\sin(\text{angle it subtends})

, so every side over the sine of its opposite angle is the same number,

2R

. Angles in a ratio therefore give sides in the ratio of their SINES, not of the angles.

Definition

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ , $R$ the circumradius. So $a : b : c = \sin A : \sin B : \sin C$ .
Angles in a ratio: find the angles first (ratio 2:3:7 of $180^\circ$ gives $30^\circ, 45^\circ, 105^\circ$ ), then take sines. Use $\sin 75^\circ = \sin 105^\circ = \frac{\sqrt3 + 1}{2\sqrt2}$ and $\sin 15^\circ = \frac{\sqrt3 - 1}{2\sqrt2}$ .
Angles in A.P. means the middle angle is $60^\circ$ .
Circumradius: $R = \dfrac{a}{2\sin A}$ . A right triangle has $R$ = half the hypotenuse.
Does the triangle exist? Given $a, b, A$ , compute $\sin B = \frac{b\sin A}{a}$ . If it exceeds 1 there is no triangle; if it is below 1, check whether both $B$ and $180^\circ - B$ leave room for $A$ .
A cevian: if $D$ divides $BC$ as $m : n$ , the sine rule in triangles $ABD$ and $ACD$ gives $\dfrac{\sin\angle BAD}{\sin\angle CAD} = \dfrac{m}{n}\cdot\dfrac{\sin B}{\sin C}$ — the $AD$ cancels.

Sine rule

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R

Rradius of the circumcircle

Worked example

The angles of a triangle are in the ratio 1 : 2 : 3. Find the ratio of its sides.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q125Moderate

Example 1 · Trigonometric Functions · Solution of Triangle — Sine, Cosine and Projection Rules

If the angles A, B and C of a triangle ABC are in the ratio 2:3:7 respectively, then the sides a, b and c are respectively in the ratio

Putting the sides in the ratio of the angles

Angles 2 : 3 : 7 do NOT give sides 2 : 3 : 7. The sides follow the sines —

\sqrt2 : 2 : (\sqrt3 + 1)

here — and the option that copies the angle ratio is the distractor.

Drill 15 more on the sine rule and the circumradius

Concept 2 of 4: The Cosine Rule — an Angle from Three Sides, a Side from Two and the Included Angle

The cosine rule is Pythagoras with a correction:

c^2 = a^2 + b^2

exactly when

C = 90^\circ

, and the

-2ab\cos C

term measures how far the angle is from a right angle. A negative cosine means an obtuse angle, and the largest angle always faces the largest side.

Definition

$c^2 = a^2 + b^2 - 2ab\cos C$ , and backwards $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$ (likewise for $A$ , $B$ ).
Use it when you have three sides, or two sides and the angle BETWEEN them — the cases the sine rule cannot start.
Largest or smallest angle: it is opposite the largest or smallest side, so compute only that one cosine. Sides 3, 5, 7: $\cos C = \frac{9 + 25 - 49}{30} = -\frac12$ , $C = 120^\circ$ .
Angles in A.P. with two sides known: $B = 60^\circ$ gives a quadratic in the third side; keep the root the stem allows.
Sides given in a ratio ( $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ ): add to get $a + b + c$ , subtract to get each side as a multiple of $k$ , then use the cosine rule.

Cosine rule

c^2=a^2+b^2-2ab\cos C \qquad \cos C=\frac{a^2+b^2-c^2}{2ab}

Worked example

Two sides of a triangle are 5 and 8 and the angle between them is

60^\circ

. Find the third side.

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q133Easy

Example 2 · Trigonometric Functions · Solution of Triangle — Sine, Cosine and Projection Rules

If the lengths of the sides of a triangle are

3, 5, 7

, then the largest angle of the triangle is

Computing the angle opposite the wrong side

The largest angle is opposite the largest side. Plugging the sides into

\cos C

in the order printed, rather than putting the largest side as

c

, gives a positive cosine and an acute answer that looks plausible.

Drill 13 more on the cosine rule — an angle from three sides, a side from two and the included angle

Concept 3 of 4: Reading an Angle off a Relation Among the Sides

Many stems never give numbers: they give an identity like

(a + b + c)(a + b - c) = 3ab

and ask for an angle. Every one of them is the cosine rule in disguise — expand, collect

a^2 + b^2 - c^2

, and divide by

2ab

Definition

Aim for $a^2 + b^2 - c^2 = k\,ab$ ; then $\cos C = \frac{k}{2}$ .
$(a + b + c)(a + b - c) = (a + b)^2 - c^2$ ; so $= 3ab$ gives $a^2 + b^2 - c^2 = ab$ , $C = 60^\circ$ ; $= ab$ would give $C = 120^\circ$ .
$2ac\cos B = c^2 + a^2 - b^2$ — watch for it disguised as $2ac\sin\frac{A - B + C}{2}$ , since $\frac{A - B + C}{2} = \frac{\pi}{2} - B$ .
Sums of cosines over sides: $\dfrac{\cos A}{a} + \dfrac{\cos B}{b} + \dfrac{\cos C}{c} = \dfrac{a^2 + b^2 + c^2}{2abc}$ .
$(a - b)^2\cos^2\frac{C}{2} + (a + b)^2\sin^2\frac{C}{2}$ expands to $a^2 + b^2 - 2ab\cos C = c^2$ .
A fourth-degree relation ( $a^4 + b^4 + c^4 = 2a^2c^2 + 2b^2c^2$ ) gives $(a^2 + b^2 - c^2)^2 = 2a^2b^2$ , so $\cos^2 C = \frac12$ : $C = 45^\circ$ or $135^\circ$ ; read the options.

Target form

a^2+b^2-c^2=k\,ab \iff \cos C=\frac{k}{2}

Worked example

\triangle ABC

(a + b)^2 - c^2 = 2ab

. Find

C

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q121Moderate

Example 3 · Trigonometric Functions · Solution of Triangle — Sine, Cosine and Projection Rules

In a triangle ABC , with usual notations,

(a + b + c)(a + b - c) = 3ab

, then

\angle C =

Losing the sign of k

a^2 + b^2 - c^2 = -ab

is an obtuse

C = 120^\circ

, not

60^\circ

. The stems that give the sum and product of two sides (

x^2 - c^2 = y

) lead exactly there.

Drill 9 more on reading an angle off a relation among the sides

Concept 4 of 4: The Projection Rule

Drop a perpendicular from

A

BC

: the foot splits

a

into

c\cos B

and

b\cos C

. That is the projection rule, and because it is LINEAR in the sides it clears identities that the quadratic cosine rule would make messy.

Definition

$a = b\cos C + c\cos B$ , $b = c\cos A + a\cos C$ , $c = a\cos B + b\cos A$ .
Sums like $(a + b)\cos C + (b + c)\cos A + (c + a)\cos B$ regroup into the three projections, giving $a + b + c$ .
$a\cos B = b\cos A$ means the two projections on $c$ are equal, so the triangle is isosceles with $a = b$ .
With a free angle: $a\cos(B - \theta) + b\cos(A + \theta) = \cos\theta(a\cos B + b\cos A) + \sin\theta(a\sin B - b\sin A) = c\cos\theta$ , because the sine rule makes the second bracket 0.

Projection rule

a=b\cos C+c\cos B \qquad b=c\cos A+a\cos C \qquad c=a\cos B+b\cos A

Worked example

Simplify

b\cos C + c\cos B + a

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q118Moderate

Example 4 · Trigonometric Functions · Solution of Triangle — Sine, Cosine and Projection Rules

(a+b)\cos C+(b+c)\cos A+(c+a)\cos B=72

and if

a=18, b=24

, then area of the triangle ABC is

Drill 4 more on the projection rule

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 (4)

The Sine Rule and the Circumradius
Sine rule
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
The Cosine Rule — an Angle from Three Sides, a Side from Two and the Included Angle
Cosine rule
$c^2=a^2+b^2-2ab\cos C \qquad \cos C=\frac{a^2+b^2-c^2}{2ab}$
Reading an Angle off a Relation Among the Sides
Target form
$a^2+b^2-c^2=k\,ab \iff \cos C=\frac{k}{2}$
The Projection Rule
Projection rule
$a=b\cos C+c\cos B \qquad b=c\cos A+a\cos C \qquad c=a\cos B+b\cos A$

Watch out for (3)

Putting the sides in the ratio of the angles
→ The Sine Rule and the Circumradius
Computing the angle opposite the wrong side
→ The Cosine Rule — an Angle from Three Sides, a Side from Two and the Included Angle
Losing the sign of k
→ Reading an Angle off a Relation Among the Sides

Test yourself on Trigonometric Functions

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Trigonometric Functions chapter test Drill all 45 questions