The Trigonometry of the Triangle

The sine and cos formula; areas of triangles and an analysis of the varios circles associated with Triangles

Contents

The Trigonometrical Formula Associated With Triangles.
The Sine Formula.
Two Additional Formulae For The Solution Of Triangles
Half Angle Formula
The Median And Centre Of Gravity ( By Apollonius )
The Orthocentre
The Angle Bisector
The Pedal Triangle
The Circumcircle
The Ex-circles.
The Triangle Formed By The Three Ex-centres
Page Comments

The Trigonometrical Formula Associated With Triangles.

There are a number of equations associated with triangles.
Of these, the best known are the Sine and Cos formulae.

The Sine Formula.

Consider the Triangle $\inline ABC$ with its Circumcircle.

A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments. A triangle with vertices $\inline A$ , $\inline B$ , and $\inline C$ is denoted $\inline \triangle ABC$ .

Acute Triangle

Draw the diameter $\inline BX$ through $\inline B$

Angle $\inline BAX = 90$ degrees and $\inline angle AXB = angle ACB$

From the diagram it can be seen that $\inline c = 2R sin C$

$\therefore\;\;\;\;\;\frac{c}{sin\,C} = 2R$

Therefore by symmetry:

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

Obtuse Triangle

Triangles can be classified according to the relative lengths of their sides:

Equilateral triangle all sides have the same length

Isosceles triangle, two sides are equal in length

Scalene triangle, all sides are unequal

If $\inline A$ is obtuse, angle $\inline BXC = 180 - A$

$a = 2R\,sin\,(180^0 - A) = 2R\,sin\,A$

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

Acute Triangle

$\inline ABC$ is an acute-angled triangle of height $\inline h$

Then $\inline CD = b\;cos\,C$

therefore $\inline BD = a - b\;cos\,C$

Using Pythagoras:

$h^2 = b^2 - (b\;cos\,C)^2$

and $\inline \;\;\;\;\;\;h^2 = c^2 - (a - b\;cos\,C)^2$

therefore $\inline b^2 - b^2\,cos^2\,C = c^2 - a^2 - b^2\;cos^2\,C + 2ab\;cos\,C$

or $\inline \;\;\;\;\;\;\;\c^2 = a^2 + b^2 - 2ab\;cos\,C}$

NOTE This equation can be re-written in terms of either angle $\inline A$ or $\inline B$

Obtuse Triangle

Triangles can also be classified according to their internal angles :

A right triangle, has one of its interior angles measuring $\inline 90^o$

A triangle that has one angle that measures more than $\inline 90^o$ is an obtuse triangle

A triangle that has all interior angles measuring less than $\inline 90^o$ is an acute triangle

If $\inline C$ is obtuse

$CD = b\;cos\,(180^0 - C)$

$BD = a + b\;cos\,(180^0 - C)$

$\therefore\;\;\;\;\;h^2 = b^2 - b^2\;cos^2\,(180^0 - C)$

and

$\;\;\;\;\;h^2 = c^2 - \left(a + b\;cos\,[180^0 - C] \right)^2$

$b^2 - b^2\;cos^2(180^0 - C)\;= c^2 - a^2 - b^2\;cos^2\,(180^0 - C)$

$- 2ab\;cos\,(180^0 - C)$

$\;\;\;\;\;c^2 = a^2 + b^2 + 2ab\;cos\,(180^0 - C)$

Thus

$\;\;\;\;\;\c^2 = a^2 + b^2 - 2ab\;cos\,C}$

Note It should be noted that the same equation can be applied in both cases.

Two Additional Formulae For The Solution Of Triangles

The cos and sine formula together are sufficient to solve any triangle but the cos formula can be unwieldy in use and is sometimes replaced by the following:

Formula 1

Using the sine formula

$\frac{a - b}{a + b} = \frac{2R\;sin\,A - 2R\;sin\,B}{2R\;sin\,A + 2R\;sin\,B}= \frac{2\;cos\,\frac{1}{2}(A + B)\times sin\;\frac{1}{2}(A - B)}{2\;sin\,\frac{1}{2}(A + B)\times cos\,\frac{1}{2}\,(A - B)}$

$= \frac{tan\,\frac{1}{2}(A - B)}{tan\,\frac{1}{2}(A + B)} = \frac{tan\,\frac{1}{2}(A - B)}{cot\,\frac{1}{2}C}$

as $\inline C$ and $\inline (A + B)$ are complements

$\therefore\;\;\;\;\;\frac{a - b}{a + b}\times cot\,\frac{C}{2} = tan\,\frac{A - B}{2}}$

This is the quickest way of solving a triangle given two sides and the included angle.

Example:

Example - Example 1

Problem

If $\inline a = 18.4$ ; $\inline b = 12.2$ and $\inline \angle C$ is $\inline 42^o$ , find the other side and the other angles.

Workings

$tan\frac{A\;-\;B}{2}\;=\;\frac{18.4\;-\;12.2}{18.4\;+\;12.2}\;cot\frac{42}{2}\;=\;\frac{6.2}{30.6}\;cot\,21$

$\therefore\;\;\;\;\;\frac{A\;-\;B}{2}\;=\;27^0\,50'$

Thus

$\;\;\;\;\;A\;-\;B\;=\;55^0\;\;40'$

But

$\;\;\;\;\;A\;+\;B\;=\;135^0$

From which

$\;\;\;\;\;A\;=\;96^0\;\;50'\;\;\;and\;\;\;B\;=\;41^0\;\;10'$

From the sine formula

$\frac{c}{Sin\,C}\;=\;\frac{b}{sin\,B}$

Solution

$\therefore\;\;\;\;\;\;c\;=\;12.2\times \frac{sin\;42^0}{sin\;41^0\;\;10'}\;=\;12.4$

Formula 2

$c = AN + NB$

$\therefore\;\;\;\;\;\f{c = a\;cos\,B + b\;cos\,A}$

Half Angle Formula

The cos formula can be used to find the ratios of the half angles in terms of the sides of the triangle and these are often used for the solution of triangles, being easier to handle than the cos formula when all three sides are given.

$cos\: C\; = \frac{a^2\;+\:b^2 - c^2}{2ab}$

but $\inline \displaystyle\;cos\;2\theta = cos^2\,\theta - sin^2\;\theta = 2\;cos^2\;\theta - 1$

Let $\inline \displaystyle C = \frac{\theta }{2}$

$\therefore\;\;\;\;\;\;2\;cos^2\,\frac{C}{2} - 1 = \frac{a^2 + b^2\;-c^2}{2ab}$

$\;\;\;\;\;\;2\;cos^2\,\frac{C}{2} = \frac{a^2 + b^2\;-c^2}{2ab}\;+1 = \frac{(a^2 + 2ab + b^2) - c^2}{2ab} = \frac{2s\times 2(s - c)}{2ab}\;\;\;\;$

where $\inline s$ is the semi perimeter of the triangle

$\therefore\;\;\;\;\;\;\cos\;\frac{C}{2} = \sqrt{\frac{s(s - c)}{ab}}$

Similarly

$1 - 2\;sin^2\,\frac{C}{2} = \frac{a^2 + b^2 - c^2}{2ab}$

$\therefore\;\;\;\;\;2\;sin^2\,\frac{C}{2} = 1 - \frac{a^2 + b^2 - c^2}{2ab}$

$= \frac{c^2\;-(a^2 - 2ab + b^2)}{2ab}= \frac{(c + a - b)(c - a + b)}{2ab}= \frac{2(s - b)(\times 2(s - a)}{2ab}$

$\therefore\;\;\;\;\;\;\sin\,\frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{ab}}$

By division

$\;\;\;\;\;\;\tan\,\frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{s(s - c)}}$

Area Of A Triangle

Let $\inline \triangle ABC$ be a triangle

The area of a triangle is a half base times height.

Various Triangle Area Formulas

$i.e.\;\;\;\;\;\;Area = \frac{1}{2}\;BC\times AD$

(1)

But

$AD = b\;sin\;C$

(2)

$S= \frac{1}{2}\,ab\,sinC = \frac{1}{2}\,bc\,sinA = \frac{1}{2}\,ac\,sin\,B$

(3)

Since the area

$S = \frac{1}{2}\;ab\;sin\,C$

(4)

The area of the triangle can be written as :

$S = \frac{1}{2}\;ab\times 2\;sin\,\frac{1}{2}C\times cos\,\frac{1}{2}\,C=$

$=ab\;\sqrt{\frac{(s - a)(s - b)}{ab}}\sqrt{\frac{s(s - c)}{ab}}$

$S = \sqrt{s(s - a)(s - b)(s - c)}$

(5)

The Median And Centre Of Gravity ( By Apollonius )

$AB^2 + AC^2 = 2AA'^2 + 2BA'^2$

$\therefore\;\;\;\;\;\;b^2 + c^2 = 2AA'^2 + 2\left(\frac{1}{2}\;a^2 \right)$

$\therefore\;\;\;\;\;4AA'^2 = 2b^2 + 2c^2\;-a^2$

$\therefore\;\;\;\;\;\;AA' = \frac{1}{2}\sqrt{2b^2 + 2c^2\;-a^2}}$

Also

$\;\;\;\;\;\;AG = \frac{2}{3}AA' = \frac{\sqrt{2b^2 + 2c^2\;-a^2}}{3}$

The Orthocentre

Let $\inline \triangle ABC$ be a regular triangle and let $\inline H$ be the orthocentre of the triangle
Using the sine formula for the triangle $\inline AHB$

$\frac{AH}{sin\;ABH} = \frac{c}{Sin\;AHB}$

But $\inline \displaystyle \angle\;ABH = (90^0 - A)$ and $\inline \displaystyle \angle\;AHB = (180^0 - C)$

$\therefore\;\;\;\;\;\frac{AH}{cos\,A} = \frac{c}{sin\,C} = 2R$

$\therefore\;\;\;\;\;\;AH = 2R\;cos\,A}$

Similarly

$\;\;\;\;\;BH = 2R\;cos\,B$

But

$\;\;\;\;\;\angle\;HBD = 90^0 - C$

Therefore

$HD = 2R\;cos\,B\;cos\,C}$

The Angle Bisector

Let $\inline \triangle ABC$ and $\inline X$ such as $\inline AX$ is the bisector of the angle $\inline \angle BAC$

As $\inline AX$ bisects the angle $\inline BAC$ internally

But $\inline \displaystyle Sin\,A = sin\,\frac{1}{2}\,A\;cos\,\frac{1}{2}\,A$

$\frac{BX}{XC} = \frac{c}{b}$

And since $\inline \displaystyle BX + XC = a$

$BX = \frac{c\,a}{b + c}$

Applying the sine formula to $\inline AXB$

$\frac{AX}{sin\,B} = \frac{BX}{sin\,\frac{1}{2}A}= \frac{a\;c}{(b + c)\;sin\,\frac{1}{2}A}$

$AX = \frac{a\;c\;sin\,B}{(b + c)\;sin\,\frac{1}{2}A}= \frac{c\;b\;sin\,A}{(b + c)\;sin\,\frac{1}{2}A}$

Therefore
The Angle bisector $\inline \displaystyle AX = \frac{2\;cb\,cos\,\frac{1}{2}A}{b + c}}$

The Pedal Triangle

The Pedal Triangle of ABC is the triangle formed by joining the feet of the altitudes of the triangle ABC.

If All The Angles Are Acute

Since $\inline FHDB$ is cyclic

$H\hat{D}F = H\hat{B}F = 90^0 - A$

And Since $\inline DHEC$ is cyclic

$F\hat{D}E = H\hat{C}E = 90^0 - A$

By addition

$F\hat{D}E = 180^0 - 2A$

$\inline H$ is the incentre of the Pedal Triangle and the angles are given by:

$(180^0 - 2A);(180^0 - 2B);(180^0 - 2C)$

Note The sides of the Pedal Triangle are $\inline a cos A$ ; $\inline b cos B$ ; and $\inline c cos C$

If A Is Obtuse

Since $\inline DBEA$ is cyclic, $\inline \angle AED = \angle ABD = B$

Since $\inline BEFC$ is cyclic, $\inline \angle FEA = \angle FBC = B$

And therefore $\inline \angle DEF = 2B$

Similarly $\inline \angle EFD = 2C$

By subtraction

$E\hat{D}F = 180^0 - 2B - 2C = 180^0 - 2(180^0 - A) = 2A - 180^0$

$\inline A$ is the Incentre of the Triangle $\inline \triangle HBC$ and the angles of the Pedal Triangle are:

$(2A - 180^0)\,;\,(2B)\,;(2C)$

Using the sine formula for triangle

$\inline AFE$ :

$\frac{EF}{sin\,EAF} = \frac{AF}{sin\,AEF}$

Hence

$\frac{EF}{sin\,EAF} = \frac{b\,cos\;(180 - A)}{sin\,HBC}$

Thus

$EF\;= - 2R\;cos\,A\;sin\,A = - a\;cos\,A$

$- R\;sin\,2A$

The sides of the Pedal Triangle are $\inline a cos A$ ; $\inline b cos B$ ; $\inline c cos C$

Note It is worth knowing that in the case of either an acute or an obtuse angle triangle, the four points $\inline A,B,C$ and $\inline H$ are the three ex-centre and incentre of the Pedal Triangle.

The Circumcircle

The radius of the circum-circle can be obtained from:

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

from which it is possible to write:

$\frac{abc}{bc\;sin\,A} = 2R$

$\frac{abc}{2\times S_{ABC}} = 2R$

$R = \frac{a\,b\,c}{4\times S_{ABC}}$

The Incircle

The Area of the triangle $\inline \triangle ABC$ is the sum of the areas of te triangles $\inline AIB$ ; $\inline AIC$ ; $\inline BIC$ .

$S_{AIB} = \frac{1}{2}\times AB\times Radius = \frac{1}{2}\;c\;r$

Similar equations can be written for triangles $\inline AIC$ and $\inline BIC$

Therefore the area of triangle $\inline ABC$ is given by:

$S_{ABC}= r\times \frac{1}{2}(a + b + c) = r\times s$

where $\inline s$ is the semi perimeter
If $\inline X , Y , Z$ are the points of contact between the triangle and circle, then

$\inline BX = BZ$ ; $\inline CX = CY$ ; $\inline AY = AZ$ and the semi circumference of the triangle( $\inline s$ ) is given by:

$BX + XC + AY = s = AY + BC$

but

$BC = a$

$\therefore\;\;\;\;\;\;\AY = s - a$

A similar relationship exists for $\inline AZ$ ; $\inline BZ$ etc.

For the triangle $\inline AZI$

$r = AI \;sin\frac{1}{2}A$

Applying the sine formula to triangle $\inline AIB$

$\frac{AI}{sin\;\frac{1}{2}B} = \frac{c}{sin\,AIB} = \frac{c}{sin\,\left(180^0 - \frac{A + B}{2} \right)}$

$= \frac{c}{sin\,\frac{1}{2}\,(A + B)} = \frac{c}{sin\,\frac{1}{2}\,C}$

$= \frac{2R\;sin\,C}{cos\,\frac{1}{2}\,C} = 4R\;sin\,\frac{1}{2}\,C$

$\therefore\;\;\;\;\;AI = 4R\;sin\,\frac{1}{2}\,B\times sin\,\frac{1}{2}\,C$

Thus

$r = 4R\;sin\,\frac{1}{2}\,A\,\times sin\,\frac{1}{2}\, B\times sin\,\frac{1}{2}\,C$

The Ex-circles.

The diagram shows the ex-circle opposite to $\inline \angle A$ .
There are of course two more circles opposite $\inline B$ and $\inline C$ . There are similar equations for them .

Let $\inline P$ , $\inline Q$ , $\inline R$ , be the points of contact between the lines which make up the sides of the triangle $\inline ABC$ and the circle.

Equating the areas of triangles $\inline ABC$ , $\inline AIB$ , $\inline AIC$ , $\inline BIC$ . we get:

$\Delta ABC = \Delta AIB + \Delta AIC - \Delta BIC$

This can be re-written in terms of $\inline r$ the radius of the ex-circle

$\Delta ABC = \frac{1}{2}\,c\,r + \frac{1}{2}\,b\,r - \frac{1}{2}\;a\,r = \frac{1}{2}\,r\,(b + c - a)$

$2\,s = a + b + c$

$\therefore\;\;\;\;\;\;\Delta ABC = r(s - a)$

Thus the radius of the ex-circle, $\inline r$ is given by the equation:

$r = \frac{\Delta ABC}{(s - a)}$

By inspection $\inline AR = AQ$ ; $\inline BR = BP$ ; $\inline CQ = CP$ .

$AB + BP = AC + CP = \frac{1}{2}\;perimeter = s$

$BP = s - c\;\;\;and\;\;\;PC = s - b$

From triangle $\inline AIR$

$r = AI\;sin\,\frac{1}{2}\,A$

Using the sine formula in triangle $\inline ABI$

$\frac{AI}{sin\,ABI} = \frac{c}{sin\,AIB}$

$\frac{AI}{sin\;(90^0 + \frac{1}{2}B)} = \frac{c}{sin\;(90^0 - \frac{1}{2}B - \frac{1}{2}A)}$

$\frac{AI}{cos\,\frac{1}{2}B} = \frac{c}{sin\frac{1}{2}C} = \frac{2R\;sin\,C}{sin\frac{1}{2}C}$

$= 4R\;cos\,\frac{1}{2}C$

$AI = 4R\,cos\,\frac{1}{2}\,B\;cos\,\frac{1}{2}\,C$

$r = 4R\;sin\,\frac{1}{2}\;A\times cos\,\frac{1}{2}\,B\times cos\,\frac{1}{2}\,cos\,C$

The Triangle Formed By The Three Ex-centres

It is clearly possible to draw a triangle based upon the three ex-centres.

Since $\inline AI_2\,,\,AI_3$ are the external bisectors of the angle $\inline A$ , the line $\inline I_2 A$ is a straight line as is also the line $\inline A\,I\,I_1$ ( the Internal bisector)
As the external and internal bisectors of an angle are perpendicular, $\inline AI_1$ is perpendicular to $\inline I_2\,I_3$
Therefore the triangle $\inline ABC$ is the Pedal Triangle of $\inline I_1,I_2,I_3$ and $\inline I$ is the orthocentre.
Since

$I_1\,BC = 90^0 - \frac{1}{2}\,B$

and

$I_1CB = 90^0 - \frac{1}{2}C$

$BI_1C = \frac{1}{2}(B + C) = 90^0 - \frac{1}{2}\,A$

therefore

$BC = I_2I_3\;cos(90^0 - \frac{1}{2}A)$

Thus

$I_2I_3 = \frac{a}{sin\frac{1}{2}A} = 2R\;\frac{sin\;A}{sin\frac{1}{2}A}$

Therefore the Length of the side of the Triangle through $\inline A$ is :

$I_2I_3 = 4R\;cos\;\frac{1}{2}A$

The nine point circle of $\inline I_1I_2I_3$ will pass through the feet of its altitudes $\inline ABC$ . The radius of the its nine point circle is therefor $\inline R$ . But since the radius of a nine point circle is half that of the circum-circle, the radius of the circum-cirle of $\inline I_1I_2I_3$ is $\inline 2R$

$\therefore\;\;\;\;\;\;II_1\; = 2(2R)\;cos\,(90^0 - \frac{1}{2}A)$

Hence

$II_1 = 4R\;sin \,\frac{1}{2}A$

Last Modified: 29 May 11 @ 16:28 Page Rendered: 2022-03-14 15:53:19