Solution of Triangles

Q: What is Napier's analogy used for?

Napier's analogy, tan((B minus C)/2) = ((b minus c)/(b plus c)) cot(A/2), solves a triangle when two sides and the included angle are known. It gives the difference of the two unknown angles, and A + B + C = 180 degrees gives their sum, so both angles follow quickly.

Q: How are the circumradius, inradius and exradii related?

The main links are r1 + r2 + r3 = 4R + r, r1 r2 + r2 r3 + r3 r1 = s squared, and 1/r1 + 1/r2 + 1/r3 = 1/r. Also r = 4R sin(A/2) sin(B/2) sin(C/2), which shows R is at least 2r, with equality only for an equilateral triangle.

Q: What is the m-n rule in a triangle?

If a point D divides BC in the ratio m : n and theta is angle ADC, then (m + n) cot theta = m cot alpha minus n cot beta = n cot B minus m cot C, where alpha and beta are the parts of angle A. It is the quickest tool for medians and other cevians.

Q: How do you find the area of a triangle from its three sides?

Use Heron's formula: area = square root of s(s minus a)(s minus b)(s minus c), where s is half the perimeter. For sides 13, 14 and 15, s = 21 and the area is 84. Once the area is known, R = abc/(4 times area) and r = area/s follow directly.

Q: Is solution of triangles in the JEE Advanced syllabus?

Yes. The JEE Advanced trigonometry syllabus lists the relations between sides and angles of a triangle: the sine rule, cosine rule, half-angle formulas and the area of a triangle. Questions often combine these with circumradius, inradius and exradius results, so practise identities as well as numerical solving.

MathsSolution Of Triangles, Heights And DistancesFor JEE aspirants

Solution of triangles means finding the unknown sides and angles of a triangle when three independent elements, at least one of them a side, are known. The working tools are the sine rule, cosine rule, projection rule, Napier's analogy, the half-angle and area formulas, and the radii $R$ , $r$ , $r_{1}$ , $r_{2}$ , $r_{3}$ of the circles linked to the triangle. This page covers every rule, the ambiguous SSA case and special points, with 19 solved examples. Solution of triangles is part of the JEE Advanced trigonometry syllabus.

Key Formulas - Quick Reference

Sine rule: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2 R$
Cosine rule: $cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$ , that is $a^{2} = b^{2} + c^{2} - 2 b c cos A$
Projection rule: $a = b cos C + c cos B$
Napier's analogy: $tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2}$
Half angles: $sin \frac{A}{2} = \frac{( s - b ) ( s - c )}{b c}$ , $cos \frac{A}{2} = \frac{s ( s - a )}{b c}$
Area: $Δ = \frac{1}{2} b c sin A = s (s - a) (s - b) (s - c) = \frac{ab c}{4 R} = r s$
Inradius: $r = \frac{Δ}{s} = (s - a) tan \frac{A}{2} = 4 R sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$
Exradius: $r_{1} = \frac{Δ}{s - a} = s tan \frac{A}{2}$ and $r_{1} + r_{2} + r_{3} = 4 R + r$
m-n rule: $(m + n) cot θ = m cot α - n cot β = n cot B - m cot C$
Median: $m_{a} = \frac{1}{2} 2 b^{2} + 2 c^{2} - a^{2}$

1. Elements and Notation of a Triangle

In triangle $A B C$ the angles at the vertices are written $A$ , $B$ , $C$ , and the sides opposite them are $a = B C$ , $b = C A$ , $c = A B$ . The semi-perimeter is $s = \frac{a + b + c}{2}$ and the area is $Δ$ . $R$ is the radius of the circumcircle and $r$ the radius of the incircle.

Solution of a triangle. The three sides and three angles are the six elements of a triangle. When any three of them are known (except the three angles alone), the other three can be calculated. Finding them is called solving the triangle. This page deals with oblique (non-right) triangles; right triangles need only basic trigonometric ratios.

$A + B + C = π$ , so $sin (A + B) = sin C$ , $cos (A + B) = - cos C$ and $cos \frac{A + B}{2} = sin \frac{C}{2}$ .
Triangle inequality: $b + c > a$ , $c + a > b$ , $a + b > c$ . Equivalently $s - a$ , $s - b$ , $s - c$ are all positive.
The larger side lies opposite the larger angle: $a > b ⟺ A > B$ .

2. Sine Rule

In any triangle the sides are proportional to the sines of the opposite angles, and the common ratio is the diameter of the circumcircle:

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

Why it holds. Draw the diameter $B A^{'}$ of the circumcircle (Figure 1).

$∠ B C A^{'} = 9 0^{\circ}$ , because it is the angle in a semicircle.
$∠ B A^{'} C = ∠ B A C = A$ , because they are angles in the same segment. If $A$ is obtuse, $∠ B A^{'} C = π - A$ , which has the same sine.
In right triangle $B C A^{'}$ : $B C = B A^{'} sin A^{'}$ , so $a = 2 R sin A$ . The same argument gives $b = 2 R sin B$ and $c = 2 R sin C$ .

Figure 1: Draw the diameter

B A^{'}

. Since

∠ B C A^{'} = 9 0^{\circ}

and

∠ B A^{'} C = A

, we get

a = 2 R sin A

, the sine rule.

Working form. Writing $a = k sin A$ , $b = k sin B$ , $c = k sin C$ (with $k = 2 R$ ) turns any relation between sides into a relation between angles, and back. Use the sine rule when a side and its opposite angle are both known (AAS, ASA) or when $R$ appears in the question.

3. Cosine Rule

In any triangle

A B C

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

The first form is often written as a formula for a side:

a^{2} = b^{2} + c^{2} - 2 b c cos A = b^{2} + c^{2} + 2 b c cos (B + C)

The cosine rule is Pythagoras' theorem with a correction term. Use it when all three sides are known (to find angles) or when two sides and the included angle are known (to find the third side).

Exam Trick

The sign of $b^{2} + c^{2} - a^{2}$ decides angle $A$ . If $a^{2} < b^{2} + c^{2}$ , $A$ is acute; if $a^{2} = b^{2} + c^{2}$ , $A = 9 0^{\circ}$ ; if $a^{2} > b^{2} + c^{2}$ , $A$ is obtuse. Test only the largest side to classify the whole triangle.

4. Projection Rule

Each side equals the sum of the projections of the other two sides on it:

a = b cos C + c cos B, b = c cos A + a cos C, c = a cos B + b cos A

In Figure 2 the altitude $A D$ splits $B C$ into $B D = c cos B$ and $D C = b cos C$ . If $B$ or $C$ is obtuse, $D$ falls outside $B C$ and one cosine is negative, so the rule still holds.

Figure 2: The projections of

A B

and

A C

B C

add up to

B C

, so

a = b cos C + c cos B

5. Napier's Analogy (Tangent Rule)

In any triangle

A B C

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

Proof in one line. By the sine rule and the sum-to-product formulas,

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

Use Napier's analogy when two sides and the included angle are given: it gives the difference of the other two angles, and $A + B + C = π$ gives their sum.

6. Half-Angle Formulas and Area of a Triangle

6.1 Trigonometric ratios of half angles

From $2 sin^{2} \frac{A}{2} = 1 - cos A$ and the cosine rule, $1 - cos A = \frac{a ^{2} - ( b - c ) ^{2}}{2 b c} = \frac{2 ( s - b ) ( s - c )}{b c}$ . The other ratios follow the same way.

Ratio	Angle $A$	Angle $B$	Angle $C$
sine of half angle	$\frac{( s - b ) ( s - c )}{b c}$	$\frac{( s - c ) ( s - a )}{c a}$	$\frac{( s - a ) ( s - b )}{ab}$
cosine of half angle	$\frac{s ( s - a )}{b c}$	$\frac{s ( s - b )}{c a}$	$\frac{s ( s - c )}{ab}$
tangent of half angle	$\frac{( s - b ) ( s - c )}{s ( s - a )}$	$\frac{( s - c ) ( s - a )}{s ( s - b )}$	$\frac{( s - a ) ( s - b )}{s ( s - c )}$

Two compact results follow at once:

tan \frac{A}{2} = \frac{Δ}{s ( s - a )} = \frac{( s - b ) ( s - c )}{Δ}, sin A = \frac{2}{b c} s (s - a) (s - b) (s - c) = \frac{2Δ}{b c}

6.2 Area of a triangle

Δ = \frac{1}{2} b c sin A = \frac{1}{2} c a sin B = \frac{1}{2} ab sin C

Δ = s (s - a) (s - b) (s - c) = \frac{ab c}{4 R} = r s

The square-root form is Heron's formula: it needs only the three sides. Handy checks: a $3$ - $4$ - $5$ triangle has $Δ = 6$ , and a $13$ - $14$ - $15$ triangle has $Δ = 84$ .

7. The m-n Rule (Cot Theorem)

Let

D

be a point on

B C

with

B D : D C = m : n

. If

∠ B A D = α

∠ D A C = β

and

∠ A D C = θ

, then

(m + n) cot θ = m cot α - n cot β

(m + n) cot θ = n cot B - m cot C

Figure 3:

B D : D C = m : n

and

θ = ∠ A D C

. Then

(m + n) cot θ = m cot α - n cot β = n cot B - m cot C

How it is proved. Drop the perpendicular from $A$ to $B C$ and express $B D$ and $D C$ through that height and the cotangents of the angles at $D$ , $B$ and $C$ ; the height cancels. For a median ( $m = n = 1$ ) the rule becomes $2 cot θ = cot α - cot β = cot B - cot C$ .

Exam Trick

Reach for the m-n rule whenever a cevian (median, angle bisector, trisector) and angles at the vertex appear together. Always take $θ$ as the angle at $D$ on the side of $C$ ; if the figure gives $∠ A D B$ , use $θ = π - ∠ A D B$ .

8. Circles Connected with a Triangle

8.1 Circumcircle

The circle through the three vertices is the circumcircle. Its centre $O$ is where the perpendicular bisectors of the sides meet, and its radius $R$ is the circumradius.

R = \frac{a}{2 sin A} = \frac{b}{2 sin B} = \frac{c}{2 sin C} = \frac{ab c}{4Δ}

8.2 Incircle

The circle touching all three sides from inside is the incircle (inscribed circle). Its centre $I$ is where the internal angle bisectors meet, and its radius $r$ is the inradius. The tangent lengths from the vertices to the incircle are $s - a$ from $A$ , $s - b$ from $B$ and $s - c$ from $C$ .

r = \frac{Δ}{s} = (s - a) tan \frac{A}{2} = (s - b) tan \frac{B}{2} = (s - c) tan \frac{C}{2}

r = \frac{a sin \frac{B}{2} sin \frac{C}{2}}{cos \frac{A}{2}} = 4 R sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

Remark (Euler's inequality). Since

sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} \leq \frac{1}{8}

(proved in Solved Example 2),

r \leq \frac{R}{2}

, that is

R \geq 2 r

. Equality holds only for an equilateral triangle.

8.3 Escribed circles (excircles)

The circle that touches side $B C$ and the extensions of $A B$ and $A C$ is the escribed circle opposite $A$ . Its radius is $r_{1}$ ; the excircles opposite $B$ and $C$ have radii $r_{2}$ and $r_{3}$ . These are the exradii. The centre $I_{1}$ is where the internal bisector of $A$ meets the external bisectors of $B$ and $C$ .

r_{1} = \frac{Δ}{s - a} = s tan \frac{A}{2} = \frac{a cos \frac{B}{2} cos \frac{C}{2}}{cos \frac{A}{2}} = 4 R sin \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

r_{2} = \frac{Δ}{s - b} = s tan \frac{B}{2} = 4 R sin \frac{B}{2} cos \frac{C}{2} cos \frac{A}{2}

r_{3} = \frac{Δ}{s - c} = s tan \frac{C}{2} = 4 R sin \frac{C}{2} cos \frac{A}{2} cos \frac{B}{2}

Figure 4: Incircle (centre

I

) and excircle opposite

A

(centre

I_{1}

), drawn to scale for

a = 4

b = 5

c = 6

r = Δ/ s

and

r_{1} = Δ/ (s - a)

Useful relations between the radii:

$r_{1} + r_{2} + r_{3} = 4 R + r$
$r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = s^{2} = \frac{r _{1} r _{2} r _{3}}{r}$
$\frac{1}{r _{1}} + \frac{1}{r _{2}} + \frac{1}{r _{3}} = \frac{1}{r}$ and $r r_{1} r_{2} r_{3} = Δ^{2}$
The largest exradius is opposite the largest side, since $r_{1} = \frac{Δ}{s - a}$ grows as $a$ grows.

9. Lengths of Medians, Angle Bisectors and Altitudes

For the lines drawn from vertex $A$ to side $B C$ :

Line from $A$	Length	Where it comes from
Angle bisector $β_{a}$	$\frac{2 b c}{b + c} cos \frac{A}{2}$	Area of $△ A B C$ = sum of the two parts cut by the bisector
Median $m_{a}$	$\frac{1}{2} 2 b^{2} + 2 c^{2} - a^{2}$	Apollonius: $b^{2} + c^{2} = 2 m_{a}^{2} + \frac{a ^{2}}{2}$
Altitude $h_{a}$	$\frac{2Δ}{a}$	$Δ = \frac{1}{2} a h_{a}$

Adding the three median formulas gives

m_{a}^{2} + m_{b}^{2} + m_{c}^{2} = \frac{3}{4} (a^{2} + b^{2} + c^{2})

10. Orthocentre, Pedal Triangle and Special Points

10.1 Orthocentre and pedal triangle

The orthocentre $H$ is the point where the three altitudes meet. The triangle $D E F$ formed by joining the feet of the altitudes is the pedal triangle (Figure 5).

Figure 5: Altitudes meet at the orthocentre

H

; their feet form the pedal triangle

D E F

O

G

H

are collinear with

O G : G H = 1 : 2

For an acute triangle $A B C$ :

The angles of the pedal triangle are $π - 2 A$ , $π - 2 B$ , $π - 2 C$ .
Its sides are $a cos A = R sin 2 A$ , $b cos B = R sin 2 B$ , $c cos C = R sin 2 C$ , and its circumradius is $\frac{R}{2}$ .
The orthocentre $H$ of $A B C$ is the incentre of the pedal triangle.
$A H = 2 R cos A$ , $B H = 2 R cos B$ , $C H = 2 R cos C$ (Solved Example 7).
The circumradii of triangles $H B C$ , $H C A$ , $H A B$ are all equal to $R$ .

10.2 Euler line and nine-point circle

The circumcentre $O$ , centroid $G$ and orthocentre $H$ are collinear (the Euler line), and $G$ divides $O H$ in the ratio $1 : 2$ .
The circumcircle of the pedal triangle passes through the midpoints of the sides and the midpoints of $A H$ , $B H$ , $C H$ . It is the nine-point circle, of radius $\frac{R}{2}$ .
The centre of the nine-point circle is the midpoint of $O H$ .

10.3 Excentral triangle

The triangle $I_{1} I_{2} I_{3}$ formed by the three excentres is the excentral triangle.

$A B C$ is the pedal triangle of $I_{1} I_{2} I_{3}$ , and the incentre $I$ of $A B C$ is the orthocentre of $I_{1} I_{2} I_{3}$ .
Its angles are $\frac{π}{2} - \frac{A}{2}$ , $\frac{π}{2} - \frac{B}{2}$ , $\frac{π}{2} - \frac{C}{2}$ .
Its sides are $I_{2} I_{3} = 4 R cos \frac{A}{2}$ , $I_{3} I_{1} = 4 R cos \frac{B}{2}$ , $I_{1} I_{2} = 4 R cos \frac{C}{2}$ .
$I I_{1} = 4 R sin \frac{A}{2} = a sec \frac{A}{2}$ , and similarly for $I I_{2}$ and $I I_{3}$ .

10.4 Distances of special points from vertex A and side BC

Point	Distance from vertex $A$	Distance from side $B C$
Circumcentre $O$	$R$	$R cos A$
Incentre $I$	$r cosec \frac{A}{2}$	$r$
Excentre $I_{1}$	$r_{1} cosec \frac{A}{2}$	$r_{1}$
Orthocentre $H$	$2 R cos A$	$2 R cos B cos C$
Centroid $G$	$\frac{1}{3} 2 b^{2} + 2 c^{2} - a^{2}$	$\frac{2Δ}{3 a}$

JEE Advanced

Distances between special points.

O H^{2} = R^{2} (1 - 8 cos A cos B cos C), O I^{2} = R^{2} - 2 R r

O G^{2} = R^{2} - \frac{a ^{2} + b ^{2} + c ^{2}}{9}

$O I^{2} = R^{2} - 2 R r \geq 0$ is another proof of $R \geq 2 r$ . The cosine formulas above assume an acute triangle; for an obtuse angle replace the cosine by its absolute value where a length is meant.

11. Regular Polygon, Sector, Segment and Cyclic Quadrilateral

11.1 Regular polygon

A regular polygon has all sides equal and all angles equal. For $n$ sides, the interior angles add up to $(n - 2) π$ , so each interior angle is $\frac{( n - 2 ) π}{n}$ . The exterior angles, taken in one direction, always add up to $2 π$ ( $36 0^{\circ}$ ). In a regular polygon the centroid, circumcentre and incentre coincide.

With side $a$ , circumradius $R$ and inradius $r$ : $a = 2 R sin \frac{π}{n} = 2 r tan \frac{π}{n}$ , and

Area = \frac{n a ^{2}}{4} cot \frac{π}{n} = \frac{n R ^{2}}{2} sin \frac{2 π}{n} = n r^{2} tan \frac{π}{n}

11.2 Sector and segment of a circle

Region	Meaning	Area ( $θ$ in radians)
Sector	Region between two radii and the arc	$\frac{1}{2} R^{2} θ$
Segment	Region between a chord and its arc	$\frac{1}{2} R^{2} (θ - sin θ)$

11.3 Cyclic quadrilateral

A cyclic quadrilateral has all four vertices on one circle. Take sides $A B = a$ , $B C = b$ , $C D = c$ , $D A = d$ and $2 s = a + b + c + d$ .

Opposite angles are supplementary: $A + C = B + D = π$ .
Ptolemy's theorem: the product of the diagonals equals the sum of the products of opposite sides, $A C \cdot B D = a c + b d$ .
$cos B = \frac{a ^{2} + b ^{2} - c ^{2} - d ^{2}}{2 ( ab + c d )}$ and $tan^{2} \frac{B}{2} = \frac{( s - a ) ( s - b )}{( s - c ) ( s - d )}$ (Solved Example 1).
Brahmagupta's formula: $Area = (s - a) (s - b) (s - c) (s - d)$ .
A circle can be inscribed in a quadrilateral if and only if the sums of opposite sides are equal: $a + c = b + d$ .

12. Solving a Triangle: Types I to V

Pick the rule from what is given. The table is the whole method in one place.

Given	Type	Rule to use	Triangles
Three sides (SSS)	I	Cosine rule or half-angle formulas	One (if the triangle inequality holds)
Two sides and the included angle (SAS)	II	Napier's analogy, or the cosine rule for the third side	One
Two sides and the difference of the opposite angles	II	Napier's analogy	One
Two angles and a side (AAS, ASA)	III	$A + B + C = π$ , then the sine rule	One
Three angles (AAA)	IV	Sine rule gives only $a : b : c = sin A : sin B : sin C$	Infinitely many (all similar)
Two sides and a non-included angle (SSA)	V	Sine rule, then check the ambiguous case	None, one or two

12.1 Type I: three sides

If the question asks for a sine or tangent of an angle, use the half-angle formulas; they need only $s$ , $s - a$ , $s - b$ , $s - c$ .
If it asks for a cosine, use the cosine rule directly: $cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$ .
Choose the form whose numbers stay simplest, then find the remaining angles and check $A + B + C = π$ .

12.2 Type II: two sides and the included angle

Given $b$ , $c$ and $A$ : Napier's analogy $tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2}$ gives $B - C$ , and $B + C = π - A$ . Solve for $B$ and $C$ , then $a = \frac{b sin A}{sin B}$ . The same analogy works when $b$ , $c$ and $B - C$ are given.

12.3 Type III and Type IV: angles given

Type III. With two angles and one side, find the third angle from $A + B + C = π$ , then the other sides from $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ .

Type IV. Three angles fix only the shape. The sine rule gives the ratio $a : b : c = sin A : sin B : sin C$ , and every triangle similar to the one found also fits.

12.4 Type V: two sides and a non-included angle (ambiguous case)

Given $b$ , $c$ and $B$ (the angle opposite $b$ ): the sine rule gives $sin C = \frac{c sin B}{b}$ , then $A = π - (B + C)$ and $a = \frac{b sin A}{sin B}$ . The catch is that $sin C$ fixes $C$ only up to $C$ or $π - C$ , so there may be no triangle, one or two. Compare $b$ with the height $c sin B$ and with $c$ (Figure 6).

Figure 6: The ambiguous case (SSA). Keep

c

and acute

B

fixed, then compare

b

with the height

c sin B

and with

c

to count the triangles.

Angle $B$	Condition on $b$	Number of triangles
Acute	$b < c sin B$	None
Acute	$b = c sin B$	One (right angle at $C$ )
Acute	$c sin B < b < c$	Two
Acute	$b \geq c$	One
Right or obtuse	$b > c$	One
Right or obtuse	$b \leq c$	None

Alternative method (quadratic in $a$ ). The cosine rule $cos B = \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c}$ rearranges to

a^{2} - (2 c cos B) a + (c^{2} - b^{2}) = 0 \Rightarrow a = c cos B \pm b^{2} - (c sin B)^{2}

Each positive root is a valid side, so counting positive roots counts triangles:

$b < c sin B$ : the square root is imaginary, so no triangle exists.
$b = c sin B$ : $a = c cos B$ . If $B$ is obtuse, $cos B < 0$ and there is no triangle; if $B$ is acute, there is exactly one.
$b > c sin B$ , $B$ acute: both roots are positive when $c cos B > b^{2} - c^{2} sin^{2} B$ , that is when $c > b$ , giving two triangles. If $b \geq c$ , only one root is positive.
$b > c sin B$ , $B$ obtuse: a positive root needs $b^{2} - c^{2} sin^{2} B > ∣ c cos B ∣$ , that is $b > c$ ; then there is one triangle. If $b \leq c$ , there is none.

Exam Trick

Two triangles appear only when the given angle is acute and the side opposite it is the shorter one, but longer than the height: $c sin B < b < c$ . In that case the two values of $C$ are supplementary, and the two third sides add up to $2 c cos B$ .

13. Solved Examples

Solved Example 1

In a cyclic quadrilateral

A B C D

with

A B = a

B C = b

C D = c

D A = d

and

2 s = a + b + c + d

, prove that

tan^{2} \frac{B}{2} = \frac{( s - a ) ( s - b )}{( s - c ) ( s - d )}

Solution:

In $△ A B C$ : $A C^{2} = a^{2} + b^{2} - 2 ab cos B$ .
In $△ A D C$ : $A C^{2} = c^{2} + d^{2} - 2 c d cos D$ . The quadrilateral is cyclic, so $D = π - B$ and $A C^{2} = c^{2} + d^{2} + 2 c d cos B$ .
Equate the two: $cos B = \frac{a ^{2} + b ^{2} - c ^{2} - d ^{2}}{2 ( ab + c d )}$ .
Then
$tan^{2} \frac{B}{2} = \frac{1 - cos B}{1 + cos B} = \frac{2 ( ab + c d ) - ( a ^{2} + b ^{2} - c ^{2} - d ^{2} )}{2 ( ab + c d ) + ( a ^{2} + b ^{2} - c ^{2} - d ^{2} )} = \frac{( c + d ) ^{2} - ( a - b ) ^{2}}{( a + b ) ^{2} - ( c - d ) ^{2}}$
Factor each difference of squares and use $2 s = a + b + c + d$ :
$tan^{2} \frac{B}{2} = \frac{( c + d - a + b ) ( c + d + a - b )}{( a + b - c + d ) ( a + b + c - d )} = \frac{2 ( s - a ) \cdot 2 ( s - b )}{2 ( s - c ) \cdot 2 ( s - d )}$

Hence $tan^{2} \frac{B}{2} = \frac{( s - a ) ( s - b )}{( s - c ) ( s - d )}$ .

Solved Example 2

In any triangle

A B C

, prove that

cos A + cos B + cos C \leq \frac{3}{2}

Solution:

$cos A + cos B = 2 cos \frac{A + B}{2} cos \frac{A - B}{2} = 2 sin \frac{C}{2} cos \frac{A - B}{2}$ and $cos C = 1 - 2 sin^{2} \frac{C}{2}$ .
Let $x = sin \frac{C}{2}$ . Then $cos A + cos B + cos C = 1 + 2 x (cos \frac{A - B}{2} - x) \leq 1 + 2 x (1 - x)$ , because $cos \frac{A - B}{2} \leq 1$ and $x > 0$ .
$x (1 - x) \leq \frac{1}{4}$ for every real $x$ , since $\frac{1}{4} - x + x^{2} = (x - \frac{1}{2})^{2} \geq 0$ .
So the sum is at most $1 + \frac{1}{2} = \frac{3}{2}$ , with equality when $A = B$ and $sin \frac{C}{2} = \frac{1}{2}$ , that is for an equilateral triangle.

Hence $cos A + cos B + cos C \leq \frac{3}{2}$ . Since the sum also equals $1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$ , this proves $sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} \leq \frac{1}{8}$ .

Solved Example 3

In a triangle,

cos A + cos B + cos C = \frac{3}{2}

. Prove that the triangle is equilateral.

Solution:

By the cosine rule: $\frac{b ^{2} + c ^{2} - a ^{2}}{2 b c} + \frac{c ^{2} + a ^{2} - b ^{2}}{2 c a} + \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab} = \frac{3}{2}$ .
Multiply by $2 ab c$ : $a (b^{2} + c^{2} - a^{2}) + b (c^{2} + a^{2} - b^{2}) + c (a^{2} + b^{2} - c^{2}) = 3 ab c$ .
Rearrange: $a (b - c)^{2} + b (c - a)^{2} + c (a - b)^{2} = a^{3} + b^{3} + c^{3} - 3 ab c$ .
Use the identity
$a^{3} + b^{3} + c^{3} - 3 ab c = \frac{a + b + c}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$
and collect terms:
$(a + b - c) (a - b)^{2} + (b + c - a) (b - c)^{2} + (c + a - b) (c - a)^{2} = 0$
Each bracket is positive by the triangle inequality and each square is at least $0$ , so every square must be $0$ .

Hence $a = b = c$ : the triangle is equilateral. (Solved Example 2 gives the same result, since equality holds only for an equilateral triangle.)

Solved Example 4

In a triangle

A B C

, prove that

\frac{a ( a + c - b )}{b ( b + c - a )} = \frac{1 - cos A}{1 - cos B}

Solution:

Multiply numerator and denominator by $a + b + c$ : $\frac{a ( a + c - b )}{b ( b + c - a )} = \frac{a [ ( a + c ) ^{2} - b ^{2} ]}{b [ ( b + c ) ^{2} - a ^{2} ]}$ .
By the cosine rule, $(a + c)^{2} - b^{2} = (a^{2} + c^{2} - b^{2}) + 2 a c = 2 a c (1 + cos B)$ and $(b + c)^{2} - a^{2} = 2 b c (1 + cos A)$ .
So the expression equals $\frac{a \cdot 2 a c ( 1 + cos B )}{b \cdot 2 b c ( 1 + cos A )} = \frac{a ^{2} ( 1 + cos B )}{b ^{2} ( 1 + cos A )}$ .
By the sine rule, $\frac{a ^{2}}{b ^{2}} = \frac{sin ^{2} A}{sin ^{2} B} = \frac{( 1 - cos A ) ( 1 + cos A )}{( 1 - cos B ) ( 1 + cos B )}$ .
Substitute and cancel $(1 + cos A) (1 + cos B)$ .

Hence $\frac{a ( a + c - b )}{b ( b + c - a )} = \frac{1 - cos A}{1 - cos B}$ .

Solved Example 5

The median

A D

of triangle

A B C

divides

∠ B A C

in the ratio

1 : 2

. Show that

\frac{sin B}{sin C} = \frac{1}{2} sec \frac{A}{3}

Solution:

$∠ B A D = A /3$ and $∠ D A C = 2 A /3$ .
Sine rule in $△ A B D$ : $\frac{B D}{sin ( A /3 )} = \frac{A D}{sin B}$ , so $A D = \frac{B D sin B}{sin ( A /3 )}$ .
Sine rule in $△ A D C$ : $A D = \frac{D C sin C}{sin ( 2 A /3 )}$ .
$A D$ is a median, so $B D = D C$ . Equating: $\frac{sin B}{sin C} = \frac{sin ( A /3 )}{sin ( 2 A /3 )} = \frac{sin ( A /3 )}{2 sin ( A /3 ) cos ( A /3 )}$ .

Hence $\frac{sin B}{sin C} = \frac{1}{2} sec \frac{A}{3}$ .

Solved Example 6

I

is the incentre of triangle

A B C

, and

R_{1}

R_{2}

R_{3}

are the circumradii of triangles

I B C

I C A

I A B

. Show that

R_{1} R_{2} R_{3} \leq R^{3}

Solution:

In $△ I B C$ , $∠ I B C = B /2$ and $∠ I C B = C /2$ , so $∠ B I C = π - \frac{B + C}{2} = \frac{π}{2} + \frac{A}{2}$ .
Sine rule in $△ I B C$ : $2 R_{1} = \frac{a}{sin ∠ B I C} = \frac{a}{cos ( A /2 )} = \frac{2 R sin A}{cos ( A /2 )} = 4 R sin \frac{A}{2}$ .
So $R_{1} = 2 R sin \frac{A}{2}$ , and similarly $R_{2} = 2 R sin \frac{B}{2}$ , $R_{3} = 2 R sin \frac{C}{2}$ .
$R_{1} R_{2} R_{3} = 8 R^{3} sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} \leq 8 R^{3} \cdot \frac{1}{8}$ , using Solved Example 2.

Hence $R_{1} R_{2} R_{3} \leq R^{3}$ , with equality for an equilateral triangle.

Solved Example 7

H

is the orthocentre of an acute triangle

A B C

. Find

A H

B H

and

C H

in terms of the angles and the circumradius

R

Solution:

$A H$ lies along the altitude from $A$ , so $∠ H A B = 9 0^{\circ} - B$ . Likewise $∠ H B A = 9 0^{\circ} - A$ .
So $∠ A H B = 18 0^{\circ} - (9 0^{\circ} - B) - (9 0^{\circ} - A) = A + B$ .
Sine rule in $△ A H B$ : $\frac{A H}{sin ( 9 0 ^{\circ} - A )} = \frac{A B}{sin ( A + B )}$ , so $A H = \frac{c cos A}{sin C}$ .
By the sine rule, $\frac{c}{sin C} = 2 R$ .

Answer: $A H = 2 R cos A$ , $B H = 2 R cos B$ , $C H = 2 R cos C$ . For an obtuse angle use $2 R ∣ cos A ∣$ .

Solved Example 8

In any triangle

A B C

, prove that

\frac{a + b}{c} = \frac{cos \frac{A - B}{2}}{sin \frac{C}{2}}

Solution:

Put $a = k sin A$ , $b = k sin B$ , $c = k sin C$ and use $sin \frac{A + B}{2} = cos \frac{C}{2}$ :

\frac{a + b}{c} = \frac{sin A + sin B}{sin C} = \frac{2 sin \frac{A + B}{2} cos \frac{A - B}{2}}{2 sin \frac{C}{2} cos \frac{C}{2}} = \frac{cos \frac{C}{2} cos \frac{A - B}{2}}{sin \frac{C}{2} cos \frac{C}{2}} = \frac{cos \frac{A - B}{2}}{sin \frac{C}{2}}

Hence proved. This is Mollweide's formula; the companion result is $\frac{a - b}{c} = sin \frac{A - B}{2} sec \frac{C}{2}$ .

Solved Example 9

In a triangle

A B C

a = 13

b = 8

and

c = 7

. Find

sin A

Solution:

Cosine rule: $cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c} = \frac{64 + 49 - 169}{2 \cdot 8 \cdot 7} = \frac{- 56}{112} = - \frac{1}{2}$ .
So $A = \frac{2 π}{3}$ (an obtuse angle, as expected since $a^{2} > b^{2} + c^{2}$ ).

Answer: $sin A = \frac{3}{2}$ .

Solved Example 10

In a triangle

A B C

, prove that

a (b cos C - c cos B) = b^{2} - c^{2}

Solution:

Method 1 (cosine rule). $ab cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2}$ and $a c cos B = \frac{a ^{2} + c ^{2} - b ^{2}}{2}$ . Subtracting, $a (b cos C - c cos B) = \frac{2 b ^{2} - 2 c ^{2}}{2} = b^{2} - c^{2}$ .

Method 2 (projection rule). From $b = a cos C + c cos A$ and $c = a cos B + b cos A$ : $a cos C = b - c cos A$ and $a cos B = c - b cos A$ . Then

a (b cos C - c cos B) = b (b - c cos A) - c (c - b cos A) = b^{2} - c^{2}

Hence proved.

Solved Example 11

In a triangle

A B C

∠ A = 6 0^{\circ}

. Find the value of

(1 + \frac{a}{c} + \frac{b}{c}) (1 + \frac{c}{b} - \frac{a}{b})

Solution:

The product is $\frac{c + a + b}{c} \cdot \frac{b + c - a}{b} = \frac{( b + c ) ^{2} - a ^{2}}{b c}$ .
$= \frac{b ^{2} + c ^{2} - a ^{2}}{b c} + 2 = 2 cos A + 2$ .
With $A = 6 0^{\circ}$ : $2 \cdot \frac{1}{2} + 2 = 3$ .

Answer: 3

Solved Example 12

Solve triangle

A B C

in which

a = 3 + 1

b = 3 - 1

and

C = 6 0^{\circ}

Solution:

$A + B = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
Napier's analogy: $tan \frac{A - B}{2} = \frac{a - b}{a + b} cot \frac{C}{2} = \frac{2}{2 3} \cdot 3 = 1$ , so $\frac{A - B}{2} = 4 5^{\circ}$ and $A - B = 9 0^{\circ}$ .
Hence $A = 10 5^{\circ}$ and $B = 1 5^{\circ}$ .
Sine rule with $sin 10 5^{\circ} = \frac{3 + 1}{2 2}$ : $c = \frac{a sin C}{sin A} = (3 + 1) \cdot \frac{3}{2} \cdot \frac{2 2}{3 + 1} = 6$ .
Check with the cosine rule: $c^{2} = (4 + 23) + (4 - 23) - 2 = 6$ .

Answer: $A = 10 5^{\circ}$ , $B = 1 5^{\circ}$ , $c = 6$ .

Solved Example 13

In a triangle

A B C

, the sides

a

b

c

are in A.P. Find

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

Solution:

$tan \frac{A}{2} = \frac{Δ}{s ( s - a )}$ and $tan \frac{C}{2} = \frac{Δ}{s ( s - c )}$ , so the product is $\frac{Δ ^{2}}{s ^{2} ( s - a ) ( s - c )}$ .
With $Δ^{2} = s (s - a) (s - b) (s - c)$ , the product is $\frac{s - b}{s} = 1 - \frac{b}{s}$ .
A.P. means $2 b = a + c$ , so $s = \frac{3 b}{2}$ and $\frac{b}{s} = \frac{2}{3}$ .

Answer: $tan \frac{A}{2} tan \frac{C}{2} = \frac{1}{3}$

Solved Example 14

In a triangle

A B C

b sin C (b cos C + c cos B) = 42

. Find the area of the triangle.

Solution:

Projection rule: $b cos C + c cos B = a$ .
So $ab sin C = 42$ .
$Δ = \frac{1}{2} ab sin C = 21$ .

Answer: 21 square units

Solved Example 15

The median

A D

of triangle

A B C

is perpendicular to

A B

. Prove that

tan A + 2 tan B = 0

Solution:

Here $B D : D C = 1 : 1$ , $α = ∠ B A D = 9 0^{\circ}$ and $β = ∠ D A C = A - 9 0^{\circ}$ .
$θ = ∠ A D C$ is an exterior angle of $△ A B D$ , so $θ = 9 0^{\circ} + B$ .
m-n rule: $(1 + 1) cot (9 0^{\circ} + B) = 1 \cdot cot 9 0^{\circ} - 1 \cdot cot (A - 9 0^{\circ})$ .
$cot (9 0^{\circ} + B) = - tan B$ , $cot 9 0^{\circ} = 0$ and $cot (A - 9 0^{\circ}) = - tan A$ , so $- 2 tan B = tan A$ .

Hence $tan A + 2 tan B = 0$ .

Solved Example 16

In a triangle

A B C

a = 13

cm,

b = 14

cm and

c = 15

cm. Find its circumradius.

Solution:

$s = \frac{13 + 14 + 15}{2} = 21$ cm.
Heron: $Δ = 21 \cdot 8 \cdot 7 \cdot 6 = 7^{2} \cdot 4^{2} \cdot 3^{2} = 84 cm^{2}$ .
$R = \frac{ab c}{4Δ} = \frac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \frac{65}{8}$ cm.

Answer: $R = \frac{65}{8}$ cm $= 8.125$ cm

Solved Example 17

In a triangle

A B C

, prove that

r_{1} + r_{2} + r_{3} - r = 4 R = 2 a cosec A

Solution:

$r_{1} + r_{2} = Δ (\frac{1}{s - a} + \frac{1}{s - b}) = \frac{Δ c}{( s - a ) ( s - b )}$ , because $(s - a) + (s - b) = c$ .
$r_{3} - r = Δ (\frac{1}{s - c} - \frac{1}{s}) = \frac{Δ c}{s ( s - c )}$ .
Adding, and using $s (s - c) + (s - a) (s - b) = 2 s^{2} - s (a + b + c) + ab = ab$ :
$r_{1} + r_{2} + r_{3} - r = Δ c \cdot \frac{s ( s - c ) + ( s - a ) ( s - b )}{s ( s - a ) ( s - b ) ( s - c )} = \frac{Δ ab c}{Δ ^{2}} = \frac{ab c}{Δ} = 4 R$
From the sine rule, $2 R = \frac{a}{sin A}$ , so $4 R = 2 a cosec A$ .

Hence proved.

Solved Example 18

The area of triangle

A B C

96

square units and its exradii are

r_{1} = 8

r_{2} = 12

r_{3} = 24

. Find the perimeter.

Solution:

$r_{1} = \frac{Δ}{s - a}$ gives $s - a = 12$ ; likewise $s - b = 8$ and $s - c = 4$ .
Adding: $3 s - (a + b + c) = 24$ , that is $3 s - 2 s = 24$ , so $s = 24$ .
Check with Heron: $24 \cdot 12 \cdot 8 \cdot 4 = 96$ . The sides are $a = 12$ , $b = 16$ , $c = 20$ .

Answer: perimeter $= 2 s = 48$ units

Solved Example 19

x

y

z

are the distances of the vertices

A

B

C

of an acute triangle from its orthocentre. Prove that (i)

\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = \frac{ab c}{x y z}

and (ii)

x + y + z = 2 (R + r)

Solution:

$x = 2 R cos A$ and $a = 2 R sin A$ , so $\frac{a}{x} = tan A$ . Hence $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = tan A + tan B + tan C$ and $\frac{ab c}{x y z} = tan A tan B tan C$ .
Because $A + B + C = π$ , $tan A + tan B + tan C = tan A tan B tan C$ . This proves (i).
$x + y + z = 2 R (cos A + cos B + cos C) = 2 R (1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2})$ .
With $r = 4 R sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$ , this is $2 R + 2 r$ . This proves (ii).

Hence both results are proved.

Practice Questions

Prove that $a sin (\frac{A}{2} + B) = (b + c) sin \frac{A}{2}$ .Answer: Write $\frac{A}{2} + B = \frac{π}{2} + \frac{B - C}{2}$ ; then both sides equal $4 R sin \frac{A}{2} cos \frac{A}{2} cos \frac{B - C}{2}$ .
Prove that $\frac{a ^{2} sin ( B - C )}{sin B + sin C} + \frac{b ^{2} sin ( C - A )}{sin C + sin A} + \frac{c ^{2} sin ( A - B )}{sin A + sin B} = 0$ .Answer: Each term equals $4 R^{2} sin A (sin B - sin C)$ (use $sin (B + C) sin (B - C) = sin^{2} B - sin^{2} C$ ); the cyclic sum is $0$ .
Prove that $\frac{c}{a - b} = \frac{tan ( A /2 ) + tan ( B /2 )}{tan ( A /2 ) - tan ( B /2 )}$ .Answer: Both sides reduce to $\frac{cos ( C /2 )}{sin ( ( A - B ) /2 )}$ .
The sides of a triangle are $a$ , $b$ and $a^{2} + ab + b^{2}$ . Prove that the greatest angle is $12 0^{\circ}$ .Answer: The cosine of the angle opposite the longest side is $- \frac{1}{2}$ .
Prove that $a (cos B + cos C) = 2 (b + c) sin^{2} \frac{A}{2}$ .Answer: By the projection rule, $a cos B + a cos C = (c - b cos A) + (b - c cos A) = (b + c) (1 - cos A)$ .
Prove that $2 (b cos^{2} \frac{C}{2} + c cos^{2} \frac{B}{2}) = a + b + c$ .Answer: The left side is $b (1 + cos C) + c (1 + cos B) = b + c + a$ by the projection rule.
Prove that $\frac{cos B}{cos C} = \frac{c - b cos A}{b - c cos A}$ .Answer: By the projection rule, $c - b cos A = a cos B$ and $b - c cos A = a cos C$ .
Prove that $\frac{cos A}{c cos B + b cos C} + \frac{cos B}{a cos C + c cos A} + \frac{cos C}{a cos B + b cos A} = \frac{a ^{2} + b ^{2} + c ^{2}}{2 ab c}$ .Answer: The denominators are $a$ , $b$ , $c$ ; then $\sum \frac{cos A}{a} = \sum \frac{b ^{2} + c ^{2} - a ^{2}}{2 ab c}$ .
In a triangle $A B C$ , $b = 3$ , $c = 5$ and $cos (B - C) = \frac{7}{25}$ . Find $tan \frac{A}{2}$ .Answer: $\frac{1}{3}$
If $x = tan \frac{B - C}{2} tan \frac{A}{2}$ , $y = tan \frac{C - A}{2} tan \frac{B}{2}$ and $z = tan \frac{A - B}{2} tan \frac{C}{2}$ , show that $x + y + z = - x y z$ .Answer: By Napier's analogy $x = \frac{b - c}{b + c}$ , $y = \frac{c - a}{c + a}$ , $z = \frac{a - b}{a + b}$ ; expanding gives $x + y + z + x y z = 0$ .
In a triangle $A B C$ , the median to $B C$ has length $\frac{1}{11 - 6 3}$ and divides angle $A$ into $3 0^{\circ}$ and $4 5^{\circ}$ . Prove that $B C = 2$ .Answer: The m-n rule gives $cot θ = \frac{3 - 1}{2}$ for $θ = ∠ A D C$ ; the sine rule in $△ A D C$ then gives $D C = 1$ , so $B C = 2$ .
Prove that $a cot A + b cot B + c cot C = 2 (R + r)$ .Answer: $a cot A = 2 R cos A$ , and $cos A + cos B + cos C = 1 + \frac{r}{R}$ .
Prove that $4 (\frac{s}{a} - 1) (\frac{s}{b} - 1) (\frac{s}{c} - 1) = \frac{r}{R}$ .Answer: The left side is $\frac{4 ( s - a ) ( s - b ) ( s - c )}{ab c} = \frac{4 Δ ^{2} / s}{4 R Δ} = \frac{Δ}{s R}$ .
$α$ , $β$ , $γ$ are the distances of the vertices of a triangle from the corresponding points of contact with the incircle. Prove that $\frac{α β γ}{α + β + γ} = r^{2}$ .Answer: $α = s - a$ , $β = s - b$ , $γ = s - c$ , so the ratio is $\frac{Δ ^{2} / s}{s} = r^{2}$ .
Prove that $r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = s^{2}$ .Answer: $r_{1} r_{2} = \frac{Δ ^{2}}{( s - a ) ( s - b )} = s (s - c)$ ; the sum is $s (3 s - 2 s) = s^{2}$ .
Prove that $r r_{1} + r r_{2} + r r_{3} = ab + b c + c a - s^{2}$ .Answer: $r r_{1} = (s - b) (s - c)$ ; summing cyclically gives $3 s^{2} - 4 s^{2} + (ab + b c + c a)$ .
Prove that $\frac{r _{1} - r}{a} + \frac{r _{2} - r}{b} = \frac{c}{r _{3}}$ .Answer: $\frac{r _{1} - r}{a} = tan \frac{A}{2}$ ; then use $tan \frac{A}{2} + tan \frac{B}{2} = \frac{c}{s} cot \frac{C}{2}$ and $r_{3} = s tan \frac{C}{2}$ .
$A$ , $A_{1}$ , $A_{2}$ , $A_{3}$ are the areas of the incircle and the three excircles of a triangle. Prove that $\frac{1}{A} = \frac{1}{A _{1}} + \frac{1}{A _{2}} + \frac{1}{A _{3}}$ .Answer: It is $\frac{1}{r} = \frac{1}{r _{1}} + \frac{1}{r _{2}} + \frac{1}{r _{3}}$ , true because $(s - a) + (s - b) + (s - c) = s$ .
In a triangle $A B C$ , $a = 5$ cm, $b = 4$ cm, $c = 3$ cm and $G$ is the centroid. Find the circumradius of $△ G A B$ .Answer: $\frac{5 13}{12}$ cm
$I$ is the incentre of triangle $A B C$ . Prove that $I A \cdot I B \cdot I C = ab c tan \frac{A}{2} tan \frac{B}{2} tan \frac{C}{2}$ .Answer: Both sides equal $4 R r^{2}$ (use $I A = r cosec \frac{A}{2}$ and $tan \frac{A}{2} tan \frac{B}{2} tan \frac{C}{2} = \frac{r}{s}$ ).
$x$ , $y$ , $z$ are the perpendiculars from the circumcentre to the sides of an acute triangle $A B C$ . Prove that $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = \frac{ab c}{4 x y z}$ .Answer: $x = R cos A$ , so both sides equal $2 tan A tan B tan C$ .
In a triangle $A B C$ , $b = 2$ cm, $c = 3$ cm and $A = \frac{π}{6}$ . Find the distance between its circumcentre and incentre.Answer: $2 - 3$ cm ( $a = 1$ , $R = 1$ , $r = \frac{3 - 1}{2}$ , $O I^{2} = R^{2} - 2 R r$ )

Common Mistakes to Avoid

Watch out

Pairing a side with the wrong angle: $a = B C$ is opposite $A$ , not the side $A B$ .
Using $s = a + b + c$ . The semi-perimeter is half of it, and every half-angle formula breaks if $s$ is wrong.
Accepting one value of $C$ from $sin C = \frac{c sin B}{b}$ in the SSA case. Check $π - C$ too, and reject it only if $B + (π - C) \geq π$ .
In Napier's analogy writing $cot \frac{B - C}{2}$ or $cot A$ . The formula is $tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2}$ .
Mixing up $r = \frac{Δ}{s}$ and $r_{1} = \frac{Δ}{s - a}$ . The exradii are always larger than $r$ .
In the m-n rule taking $θ = ∠ A D B$ instead of $∠ A D C$ , or swapping $m cot α$ and $n cot β$ .
Applying $A H = 2 R cos A$ or the pedal-triangle angles $π - 2 A$ to an obtuse triangle without adjusting signs.
Trying to find actual side lengths from three angles. Angles alone give only the ratio $a : b : c$ .
Using degrees in the sector and segment formulas $\frac{1}{2} R^{2} θ$ and $\frac{1}{2} R^{2} (θ - sin θ)$ ; $θ$ must be in radians.

Frequently Asked Questions

What is meant by solution of triangles?

Solution of triangles means finding the unknown sides and angles of a triangle from three known elements, at least one of which is a side. The sine rule, cosine rule, Napier's analogy and the half-angle formulas are used, and the choice depends on whether sides, angles or a mix are given.

When should I use the sine rule and when the cosine rule?

Use the sine rule when a side and its opposite angle are both known, as in AAS or ASA, or when the circumradius $R$ appears. Use the cosine rule when all three sides are known, or when two sides and the included angle are known and the third side is needed.

What is the ambiguous case in solution of triangles?

It is the SSA case: two sides and an angle opposite one of them. With $b$ , $c$ and acute $B$ known, there is no triangle if $b < c sin B$ , one right triangle if $b = c sin B$ , two triangles if $c sin B < b < c$ , and one triangle if $b \geq c$ .

What is Napier's analogy used for?

Napier's analogy, $tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2}$ , solves a triangle when two sides and the included angle are known. It gives the difference of the two unknown angles, and $A + B + C = π$ gives their sum, so both angles follow quickly without long cosine-rule arithmetic.

How are the circumradius, inradius and exradii related?

The main links are $r_{1} + r_{2} + r_{3} = 4 R + r$ , $r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = s^{2}$ and $\frac{1}{r _{1}} + \frac{1}{r _{2}} + \frac{1}{r _{3}} = \frac{1}{r}$ . Also $r = 4 R sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$ , which shows $R \geq 2 r$ , with equality only for an equilateral triangle.

What is the m-n rule in a triangle?

If a point $D$ divides $B C$ in the ratio $m : n$ and $θ = ∠ A D C$ , then $(m + n) cot θ = m cot α - n cot β = n cot B - m cot C$ , where $α$ and $β$ are the parts of angle $A$ . It is the quickest tool for medians and other cevians.

How do you find the area of a triangle from its three sides?

Use Heron's formula: $Δ = s (s - a) (s - b) (s - c)$ , where $s$ is half the perimeter. For sides $13$ , $14$ , $15$ , $s = 21$ and $Δ = 21 \cdot 8 \cdot 7 \cdot 6 = 84$ . Once the area is known, $R = \frac{ab c}{4Δ}$ and $r = \frac{Δ}{s}$ follow directly.

Is solution of triangles in the JEE Advanced syllabus?

Yes. The JEE Advanced trigonometry syllabus lists the relations between sides and angles of a triangle: the sine rule, cosine rule, half-angle formulas and the area of a triangle. Questions often combine these with circumradius, inradius and exradius results, so practise identities as well as numerical solving.

Previous year questions on Solution of Triangles

7 questions from past papers, each with a step-by-step solution.

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