Solution of Triangles

PROPERTIES OF TRIANGLES

In a triangle ABC, the angles are denoted by capital letters A, B, and C and the lengths of the sides opposite to these angles are denoted by small letters a, b, and c respectively. Semi-perimeter of the triangle is written as $s=\frac{a+b+c}{2},$ and its area denoted by S or $\Delta$.

$\bullet$ Sine rule: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$ , where R is the radius of the circumcircle of the $\Delta$ABC.

$\bullet$ Cosine rule: $\cos A=\frac{b^2+c^2-a^2}{2bc},\text{ cosB=}\frac{{\text{a}}^{\text{2}}+c^2-b^2}{\text{2ac}},\text{ cosC=}\frac{{\text{a}}^{\text{2}}+b^2-c^2}{\text{2ab}}.$

$\bullet$ Projection rule: a = b cosC + c cosB, b = c cosA + a cosC, c = a cosB + b cosA.

$\bullet$ Napier's analogy: $\tan \frac{B-C}{2}=\frac{b-c}{b+c}\cot \frac{A}{2},\text{ tan}\frac{\text{C-A}}{\text{2}}=\frac{c-a}{c+a}\cot \frac{B}{2},\text{ tan}\frac{\text{A-B}}{\text{2}}=\frac{a-b}{a+b}.\cot \frac{C}{2}.$

Auxiliary Formulae

$\bullet$ Trigonometric ratios of half - angles

$\sin \frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc},}\sin \frac{B}{2}=\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{ca},}\sin \frac{C}{2}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab},}$

$\text{cos}\frac{\text{A}}{\text{2}}=\sqrt{\frac{s\left(s-a\right)}{bc},}\text{ cos}\frac{\text{B}}{\text{2}}=\sqrt{\frac{s\left(s-b\right)}{ca}},\text{ cos}\frac{\text{C}}{\text{2}}=\sqrt{\frac{s\left(s-c\right)}{ab},}$

$\tan \frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}},\text{ tan}\frac{\text{B}}{\text{2}}=\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}},\text{ tan}\frac{\text{C}}{\text{2}}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}.}$

$\bullet$ Area of a triangle

$\Delta =\frac{1}{2}bc\text{ sin A = }\frac{\text{1}}{\text{2}}c\text{a sinB = }\frac{\text{1}}{\text{2}}a\text{b sin C =}\sqrt{\text{s}\left(\text{s-a}\right)\left(s-b\right)\left(s-c\right)}=\frac{abc}{4R}=rs$

where R and r are the radii of the circumcircle and the incircle of the $\Delta$ ABC

Circles Connected With Triangle

Circumcircle

The circle passing through the vertices of the triangle ABC is called the circum-circle. Its radius R is called the circum-radius. In the triangle ABC,

$R=\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}=\frac{abc}{4\Delta }$.

In-Circle

The circle touching the three sides of the triangle internally is called the inscribed or the in-circle of the triangle. Its radius r is called the in-radius of the triangle. In the triangle ABC,

$r=\frac{\Delta }{s}=\left(s-a\right)\tan \frac{A}{2}=\left(s-b\right)\tan \frac{B}{2}=\left(s-c\right)\tan \frac{C}{2}=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$

Remark:

$\bullet$ From r = 4R sin$\frac{A}{2}$ sin$\frac{B}{2}$sin$\frac{C}{2}$, we find that r $⩽$ 4R.$\frac{1}{8}$

because sin $\frac{A}{2}$ sin $\frac{B}{2}$ sin $\frac{C}{2}$ $⩽$ $\frac{1}{8}$ $\Rightarrow$ 2r $⩽$ R.

Here equality holds for the equilateral triangle.

Escribed Circles

The circle touching BC and the two sides AB and AC produced of D ABC externally is called the escribed circle opposite A. Its radius is denoted by r₁. Similarly r₂, and r₃ denote the radii of the escribed circles opposite to angles B and C respectively. r₁, r₂, r₃ are called the ex-radii of $\Delta$ ABC. Here

\begin{align}  {{r}_{1}}=\dfrac{\Delta }{s-a}=s\tan \dfrac{A}{2}=4R\sin \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}\,\,\,, \\   {{r}_{2}}=\dfrac{\Delta }{s-b}=s\tan \dfrac{B}{2}=4R\sin \dfrac{B}{2}\cos \dfrac{C}{2}\cos \dfrac{A}{2}\,\,, \\ \end{align}

$r_3=\frac{\Delta }{s-c}=s\tan \frac{C}{2}=4R\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}$,

r₁ + r₂ + r₃ = 4R + r, $r_1{r}_2+r_2{r}_3+r_3{r}_1=s^2=\frac{r_1{r}_2{r}_3}{r}$.

Regular Polygon

A regular polygon is a polygon which has all its sides as well as all its angles equal. If the polygon has 'n' sides, sum of its internal angles is (n – 2)$\pi$ and each angle is $\frac{\left(n-2\right)\pi }{n}$.

Remarks:

$\bullet$ Sum of the exterior angles of a polygon taken in one direction (clockwise or anticlockwise) remains constant and it is equal to 360°.

$\bullet$ In the regular polygon the centroid, the circumcentre, and the incentre are same.

Area of Regular Polygon

Area = $\frac{n{a}^2}{4}$ cot $\frac{\pi }{n}$ = $\frac{n{R}^2}{2}$ sin $\frac{2\pi }{n}$ = nr² tan $\frac{\pi }{n}$

(Where a is length of side, n is number of sides of polygon, R is radius of circumscribing circle and r is radius of incircle of the polygon).

Area of Sector

Area included between two radius and circumference.

Area = $\frac{R^2\theta }{2}$, where q is in radians.

Area of Segment

Area between a circumference and a chord

Area = $\frac{R^2}{2}$($\theta$ – sin $\theta$)

SOME BASIC DEFINITIONS

The Orthocentre and the Pedal Triangle

Orthocentre is a point of concurrence of altitude from the vertices to the opposite side of the triangle. The triangle LMN which is formed by joining the feet of these three altitudes is called the pedal triangle.

Remarks:

$\bullet$ Orthocentre of the triangle is the incentre of the pedal triangle.

$\bullet$ If I₁, I₂ and I₃ be the centres of escribed circles which are opposite to A, B and C respectively and I the centre of incircle then triangle ABC is the pedal triangle of the triangle I₁ I₂ I₃ and I is the orthocentre of the triangle I₁ I₂I₃ .

$\bullet$ The centroid of the triangle lies on the line joining the circumcentre to the orthocentre and divides it in the ratio 1 : 2

$\bullet$ Circle circumscribing the pedal triangle of a given triangle bisects the sides of the given triangle and also the lines joining the vertices of the given triangle to the orthocentre of the given triangle. This circle is known as nine point circle.

$\bullet$ Circum-centre of the pedal triangle of a given triangle bisects the line joining the circum-centre of the triangle to the orthocentre.

Cyclic Quadrilateral:

A cyclic Quadrilateral is a quadrilateral whose all the vertices lies on a circle.

Note:

$\bullet$Sum of the opposite angles of a cyclic quadrilateral is 180°

$\bullet$ In a cyclic quadrilateral sum of the products of the opposite sides is equal to the product of the diagonals. This is known as Ptolemy's theorem.

$\bullet$ If sum of the opposite sides of a quadrilateral is equal, then and only then a circle can be inscribed in the quadrilateral .

SOLUTION OF TRIANGLES

The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the solution of triangles.

In this chapter we will discuss the solution of oblique triangles only.

Type I: Problems based on finding the angles when three sides are given.

If the data given in sine we use the following formula which ever is applicable.

$\sin \frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc},}\sin \frac{B}{2}=\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{ca},}\sin \frac{C}{2}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab},}$

If the given data are in cosine first of all try the following formula which ever is needed.

$\text{cos}\frac{\text{A}}{\text{2}}=\sqrt{\frac{s\left(s-a\right)}{bc},}\text{ cos}\frac{\text{B}}{\text{2}}=\sqrt{\frac{s\left(s-b\right)}{ca}},\text{ cos}\frac{\text{C}}{\text{2}}=\sqrt{\frac{s\left(s-c\right)}{ab},}$

and see whether of logarithm of the number on R.H.S can be determined from the given data. If s proceed further, if not then try the following formula which ever is needed.

$\cos A=\frac{b^2+c^2-a^2}{2bc},\text{ cosB=}\frac{{\text{a}}^{\text{2}}+c^2-b^2}{\text{2ac}},\text{ cosC=}\frac{{\text{a}}^{\text{2}}+b^2-c^2}{\text{2ab}}.$

If the given data are in tangent use the following formula which ever is applicable.

$\tan \frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}},\text{ tan}\frac{\text{B}}{\text{2}}=\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}},\text{ tan}\frac{\text{C}}{\text{2}}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}.}$

Type II: Problem based on finding the angles when any two sides and the angles between them are given or any two sides and the difference of the angles opposite to them are given:

Working Rule:

Use the following formula whichever is needed.

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

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

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

Type III: Problems based on finding the sides and angles when any two angles and side opposite to one of them are given:

Working Rule:

Use the following formula whichever is needed.

(a) $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

(b) A + B + C = 180°.

Type IV: When all the three angles are given

In this case unique solution of triangle is not possible. In this case only the ratio of the sides can be determined. For this the formula.

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ can be used.

Type V: $\bullet$ If two sides b and c and the angle B (opposite to side b) are given, then$\sin C=\frac{c}{b}\sin B$,

A = 180° - (B + C) and $b=\frac{b\sin A}{\sin B}$ give the remaining elements. If b < c sin B, there is no triangle possible (fig1). If b = c sin B and B is an acute angle, then there is only one triangle possible (fig 2).If c sin B < b < c and B is an acute angle, then there are two value of angle C (fig 3). If c < b and B is an acute angle, then there is only one triangle (fig 4).

This is, sometimes, called an ambiguous case.

Alternative Method

By applying cosine rule , we have cosB =$\frac{a^2+c^2-b^2}{2ac}$

$\Rightarrow$ a² – ( 2c cosB) a + ( c² –b²) = 0

$\Rightarrow$ a = c cosB $\pm$ $\sqrt{{\left(c\cos B\right)}^2-\left(c^2-b^2\right)}$ $\Rightarrow$ a = c cosB $\pm$ $\sqrt{b^2-{\left(c\sin B\right)}^2}$

This equation leads to following cases:

Case -I If b < c sinB, no such triangle is possible.

Case -II Let b = c sin B. There are further following case:

(a) B is an obtuse angle

$\Rightarrow$ cos B is negative. There exists no such triangle.

(b) B is an acute angle

$\Rightarrow$ cos B is positive. There exists only one such triangle.

Case -III Let b > c sin B. There are further following cases:

(a) B is an acute angle

$\Rightarrow$ cos B is positive. In this case two values of a will exists if and only if

c cos B > $\sqrt{b^2-{\left(c\sin B\right)}^2}$ or c > b

$\Rightarrow$ Two such triangle is possible. If c < b, only one such triangle is possible.

(b) B is an obtuse angle

$\Rightarrow$ cos B is negative . In this case triangle will exist if and only if

$\sqrt{b^2-{\left(c\sin B\right)}^2}$ > |c cos B|

$\Rightarrow$ b > c . So in this case only one such triangle is possible.

If b < c there exists no such triangle .

$\bullet$ If one side a and angles B and C are given, then A = 180° – (B + C), and $b=\frac{a\sin B}{\sin A},c=\frac{a\sin C}{\sin A}$.

$\bullet$ If the three angles A, B, C are given, we can only find the ratios of the sides a, b, c by using sine rule (since there are infinite similar triangles possible).

Illustration 1: In a cyclic quadrilateral ABCD, prove that tan² (B/2) = $\frac{(s-a)(s-b)}{(s-c)(s-d)}$, a, b, c and d being the lengths of sides AB, BC, CD and DA respectively and 's' is semi-perimeter of quadrilateral.

Solution: In DABC

AC² = a² + b² – 2ab cos B …(i)

In DADC

AC² = c² + d² – 2cd cos D

= c² + d² – 2cd cos (180 - B)

= c² + d² + 2cd cos B …(ii)

from (i) and (ii)

cos B = $\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}$

Since tan²(B/2) = $\frac{1-\cos B}{1+\cos B}$

= $\frac{2(ab+cd)-(a^2+b^2-c^2-d^2)}{2(ab+cd)+(a^2+b^2-c^2-d^2)}$

= $\frac{{(c+d)}^2-{(a-b)}^2}{{(a+b)}^2-{(c-d)}^2}$

= $\frac{(c+d+a-b)(c+d-a+b)}{(a+b+c-d)(a+b-c+d)}$

= $\frac{(s-a)(s-b)}{(s-d)(s-c)}$ where s = a + b + c+ d

Illustration 2: In a D ABC prove that cosA + cosB + cosC $⩽$ $\frac{3}{2}$.

Solution: cosA + cosB + cosC = 2 cos $\left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}$

= 2 sin$\frac{C}{2}$ $\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right]+1$

$\left[\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\therefore \cos \left(\frac{A+B}{2}\right)=\sin \frac{C}{2}\right]$

= 4 sin $\frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$ + 1 $⩽$ 1 + $\frac{1}{8}$. 4 $\left[\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\le \frac{1}{8}\right]$ $⩽$ $\frac{3}{2}$

Illustration 3: For a triangle it is given that cos A + cos B + cos C = $\frac{3}{2}$. Prove that the triangle is equilateral.

Solution: If a, b, c are the sides of $\Delta$ ABC then given that cos A + cos B + cos C = $\frac{3}{2}$

$\Rightarrow$ $\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$

$\Rightarrow$ ab² + ac² – a³ + bc² + ba² – b³ + ca² + cb² – c³ = 3abc

$\Rightarrow$ ab² + ac² + bc² + ba² + ca² + cb² – 6abc = a³ + b³ + c³ - 3abc

$\Rightarrow$ a(b –c)² + b(c – a)² + c(a – b)² = $\frac{(a+b+c)}{2}$ [(a – b)² + (b – c)² + (c – a)²]

$\Rightarrow$ (a + b – c ) (a – b)² + [b + c – a)(b – c)² + (c + a – b)(c – a²) = 0

Now a + b > c, b + c > a, c + a > b

Since each term on the left side has positive coefficient multiplied by perfect square, each must be separately zero.

$\Rightarrow$ a = b = c. Hence, the triangle is equilateral.