MHT-CET Maths · Trigonometric Functions

Solution of a Triangle — Half-Angle Formulas, Napier's Analogy and Area

With s the semi-perimeter, the half-angle formulas express sin, cos and tan of A/2 through s and the sides; Napier's analogy links the difference of two angles to the difference of their sides; and Heron's formula gives the area from the sides alone.

All Trigonometric Functions notes MHT-CET Maths strategy

Why this matters

24 PYQs. Eleven are half-angle products and sums (tan A/2 · tan C/2, cot B/2 · cot C/2, the sum of cotangents), five are Napier's analogy, five put tan of two half-angles of a right triangle as the roots of a quadratic, and three are area. Almost every one is solved by two memorised results, so this is the fastest page of the chapter once they are in hand.

Concept 1 of 4: Half-Angle Formulas in Terms of s

The half-angle formulas trade angles for lengths. Their products are what the paper asks for, because they collapse: in

\tan\frac{A}{2}\tan\frac{C}{2}

the factors

(s - a)

and

(s - c)

cancel and only

\frac{s - b}{s}

is left. So a condition on the sides such as

a + c = 2b

turns straight into a number.

Definition

$s = \frac{a + b + c}{2}$ . $\tan\frac{A}{2} = \sqrt{\dfrac{(s - b)(s - c)}{s(s - a)}}$ , $\sin\frac{A}{2} = \sqrt{\dfrac{(s - b)(s - c)}{bc}}$ , $\cos\frac{A}{2} = \sqrt{\dfrac{s(s - a)}{bc}}$ .
Products: $\tan\frac{A}{2}\tan\frac{C}{2} = \dfrac{s - b}{s}$ , $\cot\frac{B}{2}\cot\frac{C}{2} = \dfrac{s}{s - a}$ — the leftover factor is the side NOT named.
Sides in A.P. ( $a + c = 2b$ ) means $s = \frac{3b}{2}$ , so $\tan\frac{A}{2}\tan\frac{C}{2} = \frac13$ . Conversely a product of $\frac13$ means the sides, and the angles, are in A.P.
Sum of cotangents: $\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2} = \dfrac{s^2}{\Delta}$ .
$\cos^2\frac{C}{2} = \frac{1 + \cos C}{2}$ plus the projection rule gives $a\cos^2\frac{C}{2} + c\cos^2\frac{A}{2} = s$ .
$(a + b + c)(b + c - a) = 4s(s - a) = 4bc\cos^2\frac{A}{2}$ .

Half-angle formulas

\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \qquad \tan\frac{A}{2}\tan\frac{C}{2}=\frac{s-b}{s} \qquad \cot\frac{B}{2}\cot\frac{C}{2}=\frac{s}{s-a}

ssemi-perimeter, (a + b + c)/2

Worked example

\triangle ABC

a + b = 3c

. Find

\tan\frac{A}{2}\tan\frac{B}{2}

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q144Moderate

Example 1 · Trigonometric Functions · Solution of Triangle — Half-Angle Formulas, Napier's Analogy and Area

In a triangle ABC with usual notations, if

a,b,c

are in arithmetic progression, then,

\tan\frac{A}{2}\cdot \tan\frac{C}{2}=

Using the side you were given instead of the one left over

\tan\frac{A}{2}\tan\frac{C}{2}

leaves

s - b

, not

s - a

s - c

. Name the angle that is missing from the product — its side is the one in the numerator.

Drill 10 more on half-angle formulas in terms of s

Concept 2 of 4: Napier's Analogy — the Difference of Two Angles

When a question gives two sides and the angle between them and asks about the OTHER two angles, the cosine rule is a detour. Napier's analogy goes straight to their half-difference, and

\tan\frac{B + C}{2} = \cot\frac{A}{2}

supplies the half-sum.

Definition

$\tan\dfrac{B - C}{2} = \dfrac{b - c}{b + c}\cot\dfrac{A}{2}$ (and cyclically).
Since $\frac{A + B}{2} = \frac{\pi}{2} - \frac{C}{2}$ : $\cot\frac{A + B}{2}\tan\frac{A - B}{2} = \dfrac{a - b}{a + b}$ .
Given $\cos(A - B)$ : convert to $\tan\frac{A - B}{2} = \sqrt{\frac{1 - \cos(A - B)}{1 + \cos(A - B)}}$ , apply Napier to find $\tan\frac{C}{2}$ , then $\cos C$ and the cosine rule for $c$ .
$\cot\frac{A}{2} = \frac{b + c}{a}$ means $\cos\frac{A}{2} = \cos\frac{B - C}{2}$ , so $A = B - C$ and $B = 90^\circ$ .

Napier's analogy

\tan\frac{B-C}{2}=\frac{b-c}{b+c}\cot\frac{A}{2}

Worked example

Two sides are

3

and

1

and the included angle is

60^\circ

. Find the difference of the other two angles.

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q137Hard

Example 2 · Trigonometric Functions · Solution of Triangle — Half-Angle Formulas, Napier's Analogy and Area

Two sides of a triangle are

\sqrt{3}+1

and

\sqrt{3}-1

and the included angle is

60^\circ

, then the difference of the remaining angles is

Drill 4 more on napier's analogy — the difference of two angles

Concept 3 of 4: Right Triangle: tan of the Two Half-Angles as Roots of a Quadratic

In a right triangle the two acute angles add to

90^\circ

, so their halves add to

45^\circ

and

\tan

of that sum is 1. Feed the roots' sum and product into the compound-angle formula and the coefficients of the quadratic must satisfy one fixed relation.

Definition

If $C = 90^\circ$ : $\frac{A}{2} + \frac{B}{2} = \frac{\pi}{4}$ , so $\dfrac{\tan\frac{A}{2} + \tan\frac{B}{2}}{1 - \tan\frac{A}{2}\tan\frac{B}{2}} = 1$ .
For roots of $px^2 + qx + r = 0$ : sum $-\frac{q}{p}$ , product $\frac{r}{p}$ . Then $-\frac{q}{p} = 1 - \frac{r}{p}$ , i.e. $p + q = r$ .
The letters change from paper to paper ( $a + b = c$ , $a_1 + b_1 = c_1$ ) and so does which vertex is the right angle; the relation is always 'first coefficient plus second equals the third'.

The relation

\tan\left(\tfrac{A}{2}+\tfrac{B}{2}\right)=1 \;\Rightarrow\; p+q=r \quad\text{for } px^2+qx+r=0

Worked example

In a right triangle,

\tan\frac{A}{2}

and

\tan\frac{B}{2}

are the roots of

6x^2 + qx + 1 = 0

C = 90^\circ

. Find

q

Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q126Hard

Example 3 · Trigonometric Functions · Solution of Triangle — Half-Angle Formulas, Napier's Analogy and Area

In a

\triangle PQR

m\angle R = \frac{\pi}{2}

. If

\tan\frac{P}{2}

and

\tan\frac{Q}{2}

are the roots of the equation

ax^2+bx+c=0\ (a\neq0)

, then

Drill 4 more on right triangle: tan of the two half-angles as roots of a quadratic

Concept 4 of 4: Area — Heron's Formula and Its Consequences

Once the three sides are known, the area follows without any angle: Heron's formula. And the area is the bridge back to the angles, since

\Delta = \frac12 bc\sin A

gives

\sin A = \frac{2\Delta}{bc}

Definition

$\Delta = \sqrt{s(s - a)(s - b)(s - c)}$ = $\frac12 bc\sin A$ = $\frac{abc}{4R}$ .
$\sin A = \dfrac{2\Delta}{bc}$ . The 13-14-15 triangle has $s = 21$ , $\Delta = 84$ .
Sides given through sums ( $\frac{a + b}{7} = \frac{b + c}{8} = \frac{c + a}{9} = k$ ): add to get $s = 12k$ , subtract to get $a = 4k$ , $b = 3k$ , $c = 5k$ — a right triangle, area $6k^2$ .
A side ratio that is a Pythagorean triple (5 : 12 : 13) is a right triangle: area $= \frac12$ × the two shorter sides.

Heron's formula

\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\tfrac12\,bc\sin A

Worked example

Find the area of the triangle with sides 5, 5 and 6.

Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift II · Q120Easy

Example 4 · Trigonometric Functions · Solution of Triangle — Half-Angle Formulas, Napier's Analogy and Area

In a triangle ABC with usual notations if

a = 13

b= 14,c= 15

Then

\sin A=

Drill 2 more on area — heron's formula and its consequences

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)

Half-Angle Formulas in Terms of s
Half-angle formulas
$\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \qquad \tan\frac{A}{2}\tan\frac{C}{2}=\frac{s-b}{s} \qquad \cot\frac{B}{2}\cot\frac{C}{2}=\frac{s}{s-a}$
Napier's Analogy — the Difference of Two Angles
Napier's analogy
$\tan\frac{B-C}{2}=\frac{b-c}{b+c}\cot\frac{A}{2}$
Right Triangle: tan of the Two Half-Angles as Roots of a Quadratic
The relation
$\tan\left(\tfrac{A}{2}+\tfrac{B}{2}\right)=1 \;\Rightarrow\; p+q=r \quad\text{for } px^2+qx+r=0$
Area — Heron's Formula and Its Consequences
Heron's formula
$\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\tfrac12\,bc\sin A$

Watch out for (1)

Using the side you were given instead of the one left over
→ Half-Angle Formulas in Terms of s

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