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# The Trigonometry of the Triangle

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

## 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 $\inline&space;ABC$ 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 $\inline&space;A$, $\inline&space;B$, and $\inline&space;C$ is denoted $\inline&space;\triangle&space;ABC$.

### Acute Triangle

Draw the diameter $\inline&space;BX$ through $\inline&space;B$

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

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

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

Therefore by symmetry:

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

### 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 $\inline&space;A$ is obtuse, angle $\inline&space;BXC&space;=&space;180&space;-&space;A$

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

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

### Acute Triangle

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

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

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

Using Pythagoras:

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

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

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

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

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

### Obtuse Triangle

Triangles can also be classified according to their internal angles :
• A right triangle, has one of its interior angles measuring $\inline&space;90^o$

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

• A triangle that has all interior angles measuring less than $\inline&space;90^o$ is an acute triangle
If $\inline&space;C$ is obtuse

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

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

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

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

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

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

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

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

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

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

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

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

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

Example:
##### Example - Example 1
Problem
If $\inline&space;a&space;=&space;18.4$; $\inline&space;b&space;=&space;12.2$ and $\inline&space;\angle&space;C$ is $\inline&space;42^o$ , 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
$\therefore\;\;\;\;\;\;c\;=\;12.2\times&space;\frac{sin\;42^0}{sin\;41^0\;\;10'}\;=\;12.4$

Formula 2

$c&space;=&space;AN&space;+&space;NB$

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

## 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.

$cos\:&space;C\;&space;=&space;\frac{a^2\;+\:b^2&space;-&space;c^2}{2ab}$
but $\inline&space;\displaystyle\;cos\;2\theta&space;=&space;cos^2\,\theta&space;-&space;sin^2\;\theta&space;=&space;2\;cos^2\;\theta&space;-&space;1$

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

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

or
$\;\;\;\;\;\;2\;cos^2\,\frac{C}{2}&space;=&space;\frac{a^2&space;+&space;b^2\;-c^2}{2ab}\;+1&space;=&space;\frac{(a^2&space;+&space;2ab&space;+&space;b^2)&space;-&space;c^2}{2ab}&space;=&space;\frac{2s\times&space;2(s&space;-&space;c)}{2ab}\;\;\;\;$
where $\inline&space;s$ is the semi perimeter of the triangle
$\therefore\;\;\;\;\;\;\cos\;\frac{C}{2}&space;=&space;\sqrt{\frac{s(s&space;-&space;c)}{ab}}$
Similarly

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

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

$=&space;\frac{c^2\;-(a^2&space;-&space;2ab&space;+&space;b^2)}{2ab}=&space;\frac{(c&space;+&space;a&space;-&space;b)(c&space;-&space;a&space;+&space;b)}{2ab}=&space;\frac{2(s&space;-&space;b)(\times&space;2(s&space;-&space;a)}{2ab}$
$\therefore\;\;\;\;\;\;\sin\,\frac{C}{2}&space;=&space;\sqrt{\frac{(s&space;-&space;a)(s&space;-&space;b)}{ab}}$

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

### Area Of A Triangle

Let $\inline&space;\triangle&space;ABC$ be a triangle

The area of a triangle is a half base times height.
##### Various Triangle Area Formulas
$i.e.\;\;\;\;\;\;Area&space;=&space;\frac{1}{2}\;BC\times&space;AD$

But
$AD&space;=&space;b\;sin\;C$

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

Since the area
$S&space;=&space;\frac{1}{2}\;ab\;sin\,C$

The area of the triangle can be written as :

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

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

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

## The Median And Centre Of Gravity ( By Apollonius )

$AB^2&space;+&space;AC^2&space;=&space;2AA'^2&space;+&space;2BA'^2$

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

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

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

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

## The Orthocentre

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

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

But $\inline&space;\displaystyle&space;\angle\;ABH&space;=&space;(90^0&space;-&space;A)$ and $\inline&space;\displaystyle&space;\angle\;AHB&space;=&space;(180^0&space;-&space;C)$
$\therefore\;\;\;\;\;\frac{AH}{cos\,A}&space;=&space;\frac{c}{sin\,C}&space;=&space;2R$

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

Similarly
$\;\;\;\;\;BH&space;=&space;2R\;cos\,B$

But
$\;\;\;\;\;\angle\;HBD&space;=&space;90^0&space;-&space;C$
Therefore
$HD&space;=&space;2R\;cos\,B\;cos\,C}$

## The Angle Bisector

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

As $\inline&space;AX$ bisects the angle $\inline&space;BAC$ internally

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

$\frac{BX}{XC}&space;=&space;\frac{c}{b}$

And since $\inline&space;\displaystyle&space;BX&space;+&space;XC&space;=&space;a$

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

Applying the sine formula to $\inline&space;AXB$

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

$AX&space;=&space;\frac{a\;c\;sin\,B}{(b&space;+&space;c)\;sin\,\frac{1}{2}A}=&space;\frac{c\;b\;sin\,A}{(b&space;+&space;c)\;sin\,\frac{1}{2}A}$
Therefore
The Angle bisector $\inline&space;\displaystyle&space;AX&space;=&space;\frac{2\;cb\,cos\,\frac{1}{2}A}{b&space;+&space;c}}$

## 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 $\inline&space;FHDB$ is cyclic

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

And Since $\inline&space;DHEC$ is cyclic
$F\hat{D}E&space;=&space;H\hat{C}E&space;=&space;90^0&space;-&space;A$

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

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

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

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

### If A Is Obtuse

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

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

And therefore $\inline&space;\angle&space;DEF&space;=&space;2B$

Similarly $\inline&space;\angle&space;EFD&space;=&space;2C$

By subtraction
$E\hat{D}F&space;=&space;180^0&space;-&space;2B&space;-&space;2C&space;=&space;180^0&space;-&space;2(180^0&space;-&space;A)&space;=&space;2A&space;-&space;180^0$

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

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

Using the sine formula for triangle

$\inline&space;AFE$:
$\frac{EF}{sin\,EAF}&space;=&space;\frac{AF}{sin\,AEF}$

Hence

$\frac{EF}{sin\,EAF}&space;=&space;\frac{b\,cos\;(180&space;-&space;A)}{sin\,HBC}$
Thus
$EF\;=&space;-&space;2R\;cos\,A\;sin\,A&space;=&space;-&space;a\;cos\,A$
or
$-&space;R\;sin\,2A$

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

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

## The Circumcircle

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

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

from which it is possible to write:

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

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

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

### The Incircle

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

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

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

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

$S_{ABC}=&space;r\times&space;\frac{1}{2}(a&space;+&space;b&space;+&space;c)&space;=&space;r\times&space;s$
where $\inline&space;s$ is the semi perimeter
If $\inline&space;X&space;,&space;Y&space;,&space;Z$ are the points of contact between the triangle and circle, then

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

$BX&space;+&space;XC&space;+&space;AY&space;=&space;s&space;=&space;AY&space;+&space;BC$
but
$BC&space;=&space;a$

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

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

For the triangle $\inline&space;AZI$

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

Applying the sine formula to triangle $\inline&space;AIB$

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

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

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

$\therefore\;\;\;\;\;AI&space;=&space;4R\;sin\,\frac{1}{2}\,B\times&space;sin\,\frac{1}{2}\,C$
Thus
$r&space;=&space;4R\;sin\,\frac{1}{2}\,A\,\times&space;sin\,\frac{1}{2}\,&space;B\times&space;sin\,\frac{1}{2}\,C$

## The Ex-circles.

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

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

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

$\Delta&space;ABC&space;=&space;\Delta&space;AIB&space;+&space;\Delta&space;AIC&space;-&space;\Delta&space;BIC$

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

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

$2\,s&space;=&space;a&space;+&space;b&space;+&space;c$

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

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

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

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

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

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

From triangle $\inline&space;AIR$

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

Using the sine formula in triangle $\inline&space;ABI$

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

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

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

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

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

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

## The Triangle Formed By The Three Ex-centres

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

Since $\inline&space;AI_2\,,\,AI_3$ are the external bisectors of the angle $\inline&space;A$ , the line $\inline&space;I_2&space;A$ is a straight line as is also the line $\inline&space;A\,I\,I_1$( the Internal bisector)
As the external and internal bisectors of an angle are perpendicular, $\inline&space;AI_1$ is perpendicular to $\inline&space;I_2\,I_3$
Therefore the triangle $\inline&space;ABC$ is the Pedal Triangle of $\inline&space;I_1,I_2,I_3$ and $\inline&space;I$ is the orthocentre.
Since
$I_1\,BC&space;=&space;90^0&space;-&space;\frac{1}{2}\,B$
and
$I_1CB&space;=&space;90^0&space;-&space;\frac{1}{2}C$

$BI_1C&space;=&space;\frac{1}{2}(B&space;+&space;C)&space;=&space;90^0&space;-&space;\frac{1}{2}\,A$
therefore
$BC&space;=&space;I_2I_3\;cos(90^0&space;-&space;\frac{1}{2}A)$
Thus
$I_2I_3&space;=&space;\frac{a}{sin\frac{1}{2}A}&space;=&space;2R\;\frac{sin\;A}{sin\frac{1}{2}A}$

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

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

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

$\therefore\;\;\;\;\;\;II_1\;&space;=&space;2(2R)\;cos\,(90^0&space;-&space;\frac{1}{2}A)$
Hence
$II_1&space;=&space;4R\;sin&space;\,\frac{1}{2}A$