In any triangle `ABC` , ` a/sinA= b/sin B= c/sin C` i.e., the sines of the angle are proportional to the opposite
sides.

Proof :-


Consider the acute angle triangle `AB`C
Draw `AD` perpendicular to the opposite side `BC`

In the triangle `ABD`, we have `AD/AB =sin B` , so that `AD = c sin B`

In the triangle `ACD`, we have `AD/AC = sin C` ,so that `AD = b sin C`

Equating these two values of `AD`, we have,

`c sin B= b sin C ` `b/sin B= c/ sin C`


Consider the obtuse angle `Delta ABC`

In a `Delta ABD , sin B = AD/AB = AD/c`

In a `Delta ACD , sin (pi - C)= AD/AC =AD/b`

so, `c sin B = b sin (pi - C)= b sin C`
In a similar manner, by drawing a perpendicular from `B` upon `CA`, we have


`c/sin C = a/sin A`

Consider right angled triangle `C = 90^(circ)`
If one of the angles `C`, be a right angle as in the figure, we have

`sinC=1`, `sinA =a/c` and `sin B = b/c`


So, `a/sin A= b/sinB = c/ sin C` [Because sin `C = 1`]

Hence for any type of triangle, `a/sinA = b /sin B = c/ sin C`

In a `Delta ABC`, we have `cos A = (b^2+c^2-a^2)/2b, cos B=(c^2+a^2-b^2)/2ca` and `cos C= (a^2+ b^2-c^2)/2ab`

where `a, b` & `c` are sides and `A, B` & `C` are angle of the triangle.


Proof:

Consider the acute angle `Delta ABC`,

By geometry, we have

`AB^2=BC^2+CA^2-2BC*CD` .............(i)

But `CD/CA = cos C`, so that `CD = b cos C`.

Hence (i) becomes `c^2 = a^2 + b^2 - 2ab cos C`.

`2ab cos C = a^2 + b^2 - c^2` , so `cos C =(a^2+b^2-c^2)/2ab`

Consider the obtuse angle `Delta ABC`,
By geometry, we have `AB^2 = BC^2 + CA^2 + 2BC *CD` .............(ii)

But `CD/CA =cos (angle ACD) = cos (180^(circ) - C) = -cos C`

So `CD=-b cos C`

so equation (ii) becomes

`c^2 = a^2 + b^2 + 2a ( - b cos C) = a^2 + b^2 - 2ab cos C`.

so once again, `cos C =(a^2 =b^2-c^2)/2ab`
Consider the right angle `Delta ABC`,

If `angle C=90^(circ)` , then `a^2 + b^2 = c^2`

so `cos C =(a^2 + b^2 + c^2)/2ab =(c^2-a^2)/2ab =0` , we know that `cos 90^(circ) = 0`


so here also our formula is valid. So we can say in any type of triangle `ABC, cos C = (a^2+b^2+c^2)/2ab`

similarly `cos B, cos A` can be proved.

Note :-

There is another way to prove cosine law consider the triangle as shown in figure.

`AD = AC sin C, = b sin C, CD = AC cos C = b cos C`
so `BD = BC - CD = a - b cos C`.

`Delta ADB ` is a right angle triangle so

`AB^2 = AD^2 + BD^2 , c^2 = (b sin C)^2 + (a - b cos C)^2`
`c^2 = b^2 sin^2 C + a^2 + b^2 cos^2 C - 2ab cos C = a^2 + b^2 - 2ab cos C`

`cos C= (a^2+b^2-c^2)/2ab`

Note :- (a) If the three sides of a triangle are known, we can find all the angles by using cosine rule.


(b) lf in `DeltaABC ,a> b >c`, then `angle A > angle B > angle C` or vice-versa.

In any triangle with usual notions, ` a =b cos C+ c cos B`

`b= c cos A+ a cos C`

`c = a cos B+ b cos A`

Consider the acute angle `DeltABC`


`BD/BA = cos B` so that `BD= c cos B`

an `CD/CA = cos C` so that `BD= b cos C`

Hence , `a = BC = BD + DC = c cos B + b cos C`

Consider the obtuse angle `Delta ABC`

`BD/BA = cos B` , so that `BD = c cos B` and `CD/CA = cos ACD`

`CD/CA =cos (180^(circ) -C) = - cos C` `CD = - b cos C`
Hence, in the case, `a = BC - BD = c cos B - ( - b cos C)`
`a = c cos B + b cos C`


Consider the right angle t`Delta ABC, angle C = 90^(circ)`
then `a = c cos B + 0 = c cos B + b cos 90^(circ)`

so that in each case


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

This rule is used when two sides and included angle are known.


` tan ((B-C)/2) = (b-c)/(b+c) cot (A/2)`
` tan ((C-A)/2) = (c-a)/(c+a) cot (B/2)`

` tan ((A-B)/2) = (a-b)/(a+b) cot (C/2)`

In any triangle, we have `b/c = sin B/sin C`

`:.` `(b-c)/(b+c) = (sinB- sin C) /(sin B+ sin C) = (2 cos ((B+C)/2) sin ((B-C)/2)) /(2 sin ((B+C)/2) cos ((B-C)/2))`

`=(tan ((B-C)/2) )/(tan ((B+C)/2)) )=(tan ((B-C)/2) )/(cot ((A)/2))`

`tan ((B-C)/2)= (b-c)/(b+c)* cot A/2`

`Delta =1/2 b c sin A=1/2 c a sin B= 1/2 ab sin C= sqrt (s (s-a)(s-b)(s-c))`

Different formula for area of triangle are as follows:

`AD= c sin B`

Area of triangle `ABC` is `Delta =1/2 (BC)(AD)`

`Delta =1/2 (a) (c sin B)` `=1/2 a c sin B`

`AD= b sin C, Delta =1/2 (a) (b sin C)= 1/2 ab sin C`


`Delta =1/2 ac sin B = 1/2 ab sin C = 1/2 bc sin A`
From our tenth class knowledge, we know area of triangle with sides a, b & c is denoted by `Delta`.

`Delta = 1/2 ac sin B = 1/2 ab sin C = 1/2 bc sin A`
From our tenth class knowledge, we know area of triangle with sides `a, b` & `c` is denoted by `Delta`

`Delta = sqrt (s (s-a)(s-b)(s-c))` where `s = (a+b+c)/2`

Sine of half the angles in terms of the sides . -


`sin A/2 = sqrt (((s-b)(s-c))/(bc))`
`sin B/2 = sqrt (((s-c)(s-a))/(ca))`

`sin C/2 = sqrt (((s-a)(s-b))/(ab))`


Proof:

In any triangle `ABC`, we know `cos A= (b^2 +c^2 -a^2)/2bc`


But `cos A` can be wirtten as ` cos A =1 -2 sin^2 A/2`


Hence, `2 sin^2 A/2=1 -cos A =1 -(b^2+c^2 -a^2)/2bc`

`= (2bc - (b^2+c^2)+a^2)/2b^2= (a^2 - (b^2 +c^2 -2bc))/2bc`


`= (a^2 -(b-c)^2)/2bc = ((a+b-c)(a-b+c))/2bc`


Let `2s` stand for `a + b + c`

So, `s=(a+b+c)/2` = semiperimeter.


`a + b - c = a + b + c - 2c = 2s - 2c = 2 (s - c)`
`a - b + c = a + b + c - 2b = 2s - 2c = 2 (s - b)`

so, `2 sin^2 A/2 = (2 (s-c)2(s-b))/2bc, sin A/2 = sqrt (((s-b)(s-c))/bc)`


Similarly `sin B/2 = sqrt (((s-c)(s-a))/ca)` and `sin C/2 = sqrt (((s-a)(s-b))/ab)`

` cos A/2 =sqrt ( (s (s-a))/bc)`

` cos B/2 =sqrt ( (s (s-b))/ac)`

` cos C/2 =sqrt ( (s (s-c))/ab)`


Proof:


We know `cos A = 2 cos^2 A/2 -1, 2 cos^2 A/2 =1 + cos A`

`2 cos^2 A/2 =1 + (b^2+c^2-a^2)/(2bc)= (b^2+c^2+2bc-a^2)/(2bc)= ((b+c)^2-a^2)/(2bc)`

`2 cos^2 A/2 = ((b+c+a)(b+c-a))/(2bc)`

`a + b + c = 2s` ` a + b + c - 2a = 2s - 2a = 2(s - a)`

`2 cos^2 A/2 = (2s (2)(s-a))/(2bc) = (2s (s-a))/(bc)`

`cos A/2 = sqrt ( (s (s-a))/bc)` Similarly


`cos B/2 = sqrt ( (s (s-b))/ac)`

`cos C/2 = sqrt ( (s (s-c))/ab)`

`tan A/2 = sqrt ( ((s-b)(s-c))/(s(s-a))) =(Delta/(s(s-a)))`

`tan B/2 = sqrt ( ((s-c)(s-a))/(s(s-b))) =(Delta/(s(s-b)))`

`tan C/2 = sqrt ( ((s-a)(s-b))/(s(s-c))) =(Delta/(s(s-c)))`


Proof:


since, `tan A/2 = (sin (A/2))/(cos (A/2))`

` tan A/2 = (sqrt (((s-b)(s-c))/bc))/(sqrt ((s(s-a))/bc))= sqrt( ((s-b)(s-c))/(s (s-a)))` Similarly


`tan B/2 = sqrt (((s-c)(s-a))/(s(s-b))) , tan C/2 = sqrt (((s-a)(s-b))/(s(s-c)))`

Note :-

Since, in a triangle, A is always less than `180^(circ)`, so `A/2` is always less then `90^(circ)`. Therefore, the sine, consine
A
and tangent of `A/2` (half angle) are therefore always positive.

We know `sin A =2 sin A/2 cos A/2`

But by the previous discussion

`sin A/2 = sqrt (((s-b)(s-c))/bc)` and `cos A/2 = sqrt ((s (s-a))/bc)`

So, `sin A= sqrt (((s-b)(s-c))/bc) sqrt ((s(s-a))/bc)`

`= 2/bc sqrt (s (s-a)(s-b)(s-c)) =2/bc Delta`

Let `D` be a point on the side `BC` of a `DeltaABC`, such that `BD : DC = m : n` and `angleADC = theta`,
`angle BAD =alpha` and `angle DAC = beta`. Prove that

(a) (m + n) cote theta= m cot alpha -n cot beta.
(b) (m + n) cot theta = n cot beta - m cot C.

Proof:

(a) Given `BD/DC =m/n ` and `angle ADC = theta`

`angle ADB = 180^(circ) - theta , angle BAD= alpha` and `angle DAC= beta`

So, `angle ABD = theta- alpha = B`, `C = 180^(circ)-(theta + beta)`


From `Delta ABD, BD/ (sin alpha) = AD/ (sin (theta - alpha)) .........(i)

From `Delta ADC, EC/ (sin beta ) = AD/(sin (theta+ beta)).................(ii)
Dividing equation (i) by (ii)

`(bd /dc )* (sin beta)/ (sin alpha)= sin(theta + beta)/(sin(theta - alpha)) => (m sin beta)/(n sin alpha) =(sin(theta + beta))/(sin(theta-alpha))`.............(iii)

` (m sin beta)/(n sin alpha)= ( sin theta cos beta+ cos theta sin beta)/(sin theta cos alpha -cos theta sin alpah)`

`=>m sin beta (sin theta cos alpha - cos theta sin alpha) = n sin alpha (sine theta cos beta +cos theta + sin beta)`

Now dividing both sides by `sin alpha sin beta sin theta`

`=> m cot alpha - m cot theta = n cot beta + n cot theta`

`=>(m + n) cot theta = m cot alpha - n cot beta`

(b) We have, `angleCAD= 180^(circ) - (theta+C)`

`angle ABC = B, angle ACD = C, angle BAD = (theta - B)`

Putting these values in equation (iii) we get

`m sin (theta+ C) sin B = n sin C sin (theta - B)`
`m (sin theta cos C+cos theta sin C) sin B = n sin C (sin theta cos B- cos theta sin B)`

dividing both sides by `sin theta` sin B sin C`

`=> m (cot C +cot theta) = n (cot B - cot theta)`
`(m + n) cot theta = ncot B - m cotC`

If `ABC` be any triangle, and `D, E` and `F` respectively the middle
points of `BC, CA` and `AB` the lines `AD, BE` and `CF` are called
the medians of the triangle.


`AD = 1/2 sqrt (2b^2 +2c^2-a^2)`

`BE = 1/2 sqrt (2c^2 +2a^2-b^2)`

`CF = 1/2 sqrt (2a^2+2b^2-c^2)`


From geometry, we know that the medians meet in a common
point `G,` such that

`AG=2/3AD, BG=2/3BE` and `CG=2/3 CF`

The point `G` is called the centroid of the triangle.

Length of the Medians :
`AD^2+AC^2+CD^2-2ACCD cos C=b ^2 +(a^2/4) -2b (a/2)cos C`

`AD^2=b^2+(a^2/4)- ab cos C`

and `c^2 = b^2 + a^2 - 2ab cos C`

So, `2AD^2 - c^2 = b^2 -(a^2/2)` so that `AD =1/2 sqrt (2b^2+2c^2-a^2)`

we can also write, `AD= 1/2 sqrt(b^2 +c^2+b^2+c^2-a^2)`

`AD=1/2 sqrt (b^2+c^2 +2bc cos A)` similarly

`BE =1/2 sqrt(2c^2+2a^2-b^2)` and `CF =1/2 sqrt (2a^2 +2b^2-c^2)`

Let `angle BAD= beta ` and `angle CAD = gamma`, we have

`sin gamma /sin C= DC /AD= a /2x, sin gamma =a/2x sin C`, where `AD=x (say)`


`x =AD =1/2 sqrt (2b^2+2c^2-a^2), sin gamma = (a sin C)/(sqrt (2b^2+2c^2-a^2))`


Similarly, `sin beta = (a sin B)/ (sqrt (2b^2+2c^2-a^2))`

Again, ifthe `angle ADC = theta` , we have

`(sin theta)/(sin C)= (AC) /(AD) =b/x , sin theta =b/x sin C= (2bsin C)/(sqrt (2b^2+2c^2-a^2))`


Note : The centroid lies on the line segment joining the circumcentre to the orthocentre and divides the line
segment in the ratio `1 : 2`.

Let `O` and `H` be the circumcentre and orthocentre respectively.
Draw `OD` and `HK` perpendicular to `BC`.
Let `AD` and `OH` meet in `G`. By geometry `DeltaAGP` and `Delta.OGD`
are similar

`OG/GP =AG/GD =2/1`


The centroid therefore lies on the line segment joining the circumcentre to the orthocentre and divides it
in the ratio `1 : 2`.

If `AD` bisects the angle `A` and divide the base into portions `x` and `y`, we have by geometry. The length of
bisectors will be as follows :

`AD = (2bc)/(b+c) cos A/2`

`BE= (2ca)/(c+a) cos B/2`


`CF = (2ab)/(A+b) cos C/2`

`x/y = AB/AC =c/b, a/c =y/b ` `=> (x+y)/(b+c)=(a)/(b+c)` ....................(i)


giving `x` and `y`
Also, if `delta` be the length of `AD` and `theta` the angle it makes with `BC`, we have


`Delta ABD+ Delta ACD =Delt ABC`, `1/2 c delta sin )A)/(2)+1/2 b delta sin ((A)/(2))=1/2 bc sin A`

`delta =(( bc) / (b+c) )* (sin A)/(sin (A/2))= (2bc )/(b+c) cos (A/2)`
`delta = ((bc)/(b+c))* (sin A)/(sin(A/2))= (2bc )/(b+c) cos (A/2)`

`theta = angle DAB +B =(A/2)+B`

Thus, we have the length of bisector and its inclination to `BC`.

To find the magnitude of R, the radius of the circum circle of any triangle `ABC`

`2R=a /sin A=b/sin B=c/sin C` `R =abc/(4Delta)`


Proof: Consider any triangle `ABC` as shown in three figure

Bisecting the two sides `BC` and `CA` in `D` and `E` respectivley and draw `DO` and `EO` perpendicular to `BC`
and `CA`.
By geometry, `O` is the centre of the circumcircle. Join `OB` and `OC`.
The point `O` may either lie within the triangle as in figure `(1)` or without it as in figure `(2)` or upon one of
the sides as in figure `(3 )`.

Taking the first figure, the two triangles `BOD` and `COD` are equal in all respects, so that
`angle BOD = angle COD`, `:. angleBOD = 1/2 (2 angleBAC) = angle BAC =A`

Also, `BD = BO sin (BOD)= BO sinA = R sinA` [as `R = BO`]
`a/2 = R sinA`
If `A` be obtuse, as in figure (2), we have

`angleBOD = 1/2 angle BOC= angle BLC = 180^(circ) - A`

`sin(angleBOD ) = sin ( 180^(circ)- A) = sin A`
and `R =a/2sinA`
If `A` be right angle as in figure(3) we have

`R = OA = OC = a/2 = a/(2 sin A)= sin 90^(circ) = 1`

so in all the three cases, we have
`R=a/(2sin A)=b/(2sin B) =c /( 2sin C)`


as we know `Delta = (1/2) bc sin A`

As `sinA= (a/2R)`
`:.` `Delta =1/2 bc (a/2R)=(abc)/(4R)`

`R=(abc)/(4Delta)`

Note:


(a) ln case of acute angle triangle, circumcentre lies within the triangle.
(h) ln case of obtuse angle triangle, circumcentre lies outside the triangle.
(c) In case of right angle triangle, circumcentre lies on the mid point of hypotenuse.

Radius of incircle ` r= Delta /s = (s-a)tan A/2=(s-b)tan (B/2)=(s-c) tan C/2`

`r= 4Rsin A/2sin B/2 sin C/2`

The value of `r` the radius of the incircle of the triangle `ABC` :

Proof: Consider the triangle `ABC` as shown in figure. Bisect the two
angles `B` and `C` by the two lines `BI` and `CI` meeting in `I`
By geometry, `I` is the centre of the incircle,join `IA`, and draw
`ID`, `IE` and `IF` perpendicular to the three sides.
The `ID = IE = IF = r`

We have Area of `Delta IBC =1/2 (ID)(BC)=1/2 ra`

Area of `Delta ICA=1/2 (IE)(AC) =1/2 r*b` Area of `Delta IAB =1/2 (IF)(AB)=1/2 r*c`

Hence by addition, we have


`1/2 r(a+b+c)=` sum of areas of `Delta IBC, Delta IAC , Delta IBA`

`= Delta ABC = Delta`

`r = (Delta)/ ((a+b+c)/2) =Delta/s`

The same `r` can be expressed in a different way also :
Consider the same figure `ABC` as shown above. The angles `IBD` and `IDB` are respectively equal to the
angles `IBF` and `IFB`, so the two triangle `IDB` and `IFB` are equal in all respects.


Hence, `BD = BF` , so that `2BD = BD+BF`

so also, `A E = AF` , so that `2A E = A E + AF`

and `CE = CD`, so that `2CE = CE + CD`

Hence, by addition, we have

`2BD +2 A E + 2C E = (BD+CD) +(BF+AF)+(AE+CE)`

`:.` `2BD + 2b = a + b + c = 2s`,


Hence, `BD = s - b = BF`, similarly `CE = s - c`, `AF = s - a`


tan IBD=ID /BD =r/(s-b)= tan B/2`


`r=(s-b)tan (B/2)= (s-a) tan (A/2)= (s-c)tan (C/2)`

A third value of `r` may be found as follows :

`a = BD+CD =ID cot IBD +ID cot ICD`

`= rc ot B/2 + r cot C/2 =r [ ( (cos B/2)/(sin (B/2)))+((cos (C/2))/(sin (C/2)))]`

`a = r ((sin ((B+C)/2))/(sin B/2 sin C/2))=r ((cos A/2)/(sin (B/2)sin (C/2)))`


`r= a ( (sin B/2 sin C/2)/(cos A/2))` [as ` a/(sin A)=2R`]

As `a = 2Rsin A = 2R2 Sin A/2 cos A/2`


`:.` `r= 4 R sin A/2 sin B/2 sin C/2`

To find the value of `r _1` , the radius of the escribed circle opposite the angle
A of the triangle `ABC` :


`r_1 =Delta/(s-a) =s tan A/2 = 4 R sin A/2 cos B/2 cos C/2`


`r_2 =Delta/(s-b) =s tan B/2 = 4 R sin A/2 cos B/2 cos C/2`

`r_3 =Delta/(s-c) =s tan C/2 = 4 R sin A/2 cos B/2 cos C/2`


Proof: Produce `AB` and `AC` to `L` and `M`. Bisect the angles `CBL` and `BCM` by the lines `BI_1` and `CI_1` and let these
lines meet in `I_1`
-
Draw `I_1D_1` , `I_1E_1` and `I_1F_1` perpendicular to these lines respectively.

The two triangles `I_1D_1B` an `I_1F_1B ` are equal in all respect , so that `I_1F_1= I_1D_1` similarly `I_1E_1 =I_1D_1`

The three perpendicular ` l 1_1D_1, I_1E_1` and `I_1F_1` being equal, the point `I_1` is the centre of the required circle.
Now, the area `ABI_1C` is equal to the triangles `ABC` and `I_1BC`, it is also equal to the sum of the triangle
`I_1BA` and `I_1CA`.

Hence , `DeltaABC +DeltaI_1BC= Delta I_1CA+ DeltaI_1AB`

`:.` `Delta +1/2 (I_1D_1)(BC)=1/2 (I_1E_1)(CA)+1/2 (I_1F_1)(AB)`

`:.` `Delta +1/2 r_1 a =1/2 r_1*b+1/2 r_1 c`

`Delta = r_1/2 (b+c-a)=r_1/2(a+b+c-2a)=r_1 (s-a)`

`r_1 =Delta/(s-a)` similarly `r_2= Delta/(s-b), r_3 =Delta/(s-c)`

A Second Value of `r_1` can be obtained :
Since `A E_1` and `AF_1` are tangents, ` A E_1 = AF_1`
Similarly, `BF_1 = BD_1` and `CE_1 = CD_1`

`:.` `2A E_1 = A E_1 + AF_ 1 = AB + BF_1 + AC + CE_1`
`= AB + BD_1 + AC + CD_1`
`= AB + AC + BC = 2s`

`:.` `A E_1 = s = AF_1`
Also `BD_1= BF_1 = AF_1 - AB = s - c`
similarly `CD_1 = CE_1 = A E_1 - AC = s - b` `:.` `I_1E_1=A E_1 tan (I_1 A E_1)`

so, `r_1 = s tan A/2`

A third value of `r_1` may be obtained
For, since `I_1C` bisects the angle `BCE_1`, we have

`angle I_1 CD_1 =1/2(180^(circ)-C)=90^(circ)-C/2`

so, `angle I_1BD_1=90^(circ) -B/2`

`:.` `a = BC = BD_1 + D_1C = I_1 D_ 1 cot I_1BD _1 + I_1D_ 1 cot I_1CD_1`

`= r_1 ( tan B/2 + tan C/2)= r_1 ( ( (sin B/2)/(cos B/2) )+( (sin C/2)/(cos C/2) ) )= r_1 (sin ((B+C)/2))/(cos B/2 cos C/2)`


`=> a cos B/2 cos C/2 = r_1 cos A/2`

`r_1 = ( a cos B/2 cos C/2 )/(cos A/2) ` as `a = 2Rsin A= 4Rsin A/2 cos A/2`


`:.` `r_1 4R sin A/2 cos B/2 cos C/2`

similarly , `r_2 =4R sin B/2 cos C/2 cos A/2`

`r_3 = 4R sin C/2 cos A/2 cos B/2`


Important results regarding `r _1, r_2` and `r_3`.


Givenr 'r_1, r_2`, and `r_3`


(i) semiperimeter =`s =sqrt (r_1_2+r_2r_3+r_3r_1)= sqrt (sum r_1r_2) ` (ii) `Delta =(r_1r_2r_3)/(sqrt (sum r_1r_2))`

(iii) `r= (r_1r_2r_3)/(sumr_1r_2)` (iv) `R= ((r_1+r_2)(r_2+r_3)(r_3+r_1))/(4 sum r_1r_2)`


(v) `a = (r_1 (r_2+r_3))/(sqrt(sum r_1r_2)), b = (r_2 (r_3+r_1))/(sqrt(sum r_1r_2)) , c= (r_3 (r_1+r_2))/(sqrt(sum r_1r_2))` (vi) `sin A = (2r_1 sqrt (sum r_2r_3))/( (r_1+r_2)(r_1+r_3))`

Let `ABC` be any triangle and let `AK, BL` and `CM` to be perpendiculars from `A, B` and `C` upon the opposite sides of the triangle. It can be easily shown from geometry, that these three perpendiculars meet in a common point `H`. This point `H` is called the orthocentre of the triangle. The triangle `KLM`, which is formed by joining the feet of these perpendicular, is called the pedal triangle of `ABC`.

Distances of the orthocentre of the angular points of the triangle :

Consider an acute angle triangle `ABC`.
We have, `HK = KB tan (HBK) = KB tan (90^(circ) - C) = KB cot C`


`= AB cos B cot C =AB cos B (cosC/sin C) =AB/ sin C cos B cos C`

`HK= c/ sin C cos B cos C = 2R cos B cos C`

Agaiin, `AH = AL sec (KAC)= c cos A sec (90^(circ) -C)`

`= c cos A cosec C =c / sin C cos A = 2R cos A`

similarly `BH =2R cos B `and `CH = 2R cos C`.


The distances of the orthocentre from the angular point are therefore, `2R cos A, 2R cos B` and
`2R cos C`. Its distance from the sides `a, b, c` are `2R cos B cos C, 2R cos C cos A and 2R cos A cos B`
respectively.

The sides and angles to the pedal triangle :

Consider an acute angle triangle `ABC`:
Since the angles `HKC` and `H LC` are right angles, the points `H, L, C` and `K` lie on a circle.
`angle HKL = angle HCL = 90^(circ) - A` Similarly.

`H, K, B, M` lie on a circle, and therefore,
`angle HKM = angle HBM= 90^(circ) - A`
Hence `angle MKL = 180^(circ) - 2A = the supplement of 2A`.
` angle KLM =180^(circ) - 28, angle LMK = 180^(circ) - 2C`.

Again, from the triangle `ALM`, we have

`(LM)/ sin A= (AL)/ (sin (AML))=( AB cos A)/(cos (HML))= (c cos A)/(cos (HAL)) = (c cos A)/(sin C)`

`LM= c /(sin C ) cos A sin A =c/(sin C) sin A cos A= a cos A`

so, `LM = a cos A`, similarly `MK = b cos B, KL = c cos C`

The sides of the pedal triangle therefore `a cos A, b cos Band c cos C`; also its angles are the supplements
of twice the angles of the triangle.

Let `l` be the centre of the incircle and `I_1 I_2` and `I_3` the centres of
the escribed circles. Which are opposite to `A, B` and `C`
respectively. `IC` bisects the angle `ACB` and `I_1 C` bisects the angle
`BCM`.


`:.` `angle ICI_1 = angle ICB + angle I_1CB =1/2 angle ACB +1/2 angle MCB`

`=1/2 ( angle ACB + angle MCB)=1/2 (180^(circ))=90^(circ)`

`= A` right angle.

Similarly,`angle ICI_2`, is a right angle. Hence `I_1C I_2` is a straight line to which `IC` is perpendicular. So, `I_2AI_3` is
a straight line to which `lA` is perpendiculars and `I_3BI_ 1` is a straight line to which `IB` is perpendicular.

Also, since `IA` and `l _1A` both bisect the angle `BAC`, the three points `A`, `l` and `I_1` are in a striaght line.
Similarly, `B I I_2`, and `CI I_3` are straight lines. Hence, `I_1 I_2 I_3` is a triangle, which is such that `A, B` and `C` are
the feet of the perpendiculars drawn from its vertices upon the opposite sides and such that `l` is the
intersection of these perpendicular. i.e. `ABC` is its pedal triangle and lis its orthocentre. The triangle`I_1I_2I_3` is often called the ex centric triangle.

`OH = R sqrt (1 - 8 cos A cos B cos C )`

If `OF` be perpendicular to `AB, we have
`angle OAF = 90^(circ) - angle AOF = 90^(circ) - C`
Also, `angle OAH = A- angle OAF - angle HAL`
`= A- 2 (90^(circ) - C) = A+ 2C - 180^(circ)`
`= A+2C - (A+ B +C) = C - B`
Also, `OA = R, HA = 2 RcosA`

`OH^2 = OA^2 + HA^2 - 2 OA HAcos OAP = R^2 +4R^2 cos^2 A- 4R^2 cos A cos (C - B)`

`= R^2 + 4R^2 cos A [cos A- cos (C - B)]`
`= R^2 - 4R^2 cos A [cos (B + C)+ cos (C - B)]`
`OH^2 = R^2 - 8R^2 cos A cos B cos C`

`OH =R sqrt (1-8 cos Acos B cos C)`

`OI = sqrt (R^2 -2Rr)`

Let `O` be the circumcentre and `OF` be perpendicular to `AB`.
Let `I`l be the incentre and `IE` be perpendicular to `AC`.
Then, as in the previous article,

`angle OAF = 90^(circ) -C`

`:.` `angle OAI = angle IAF - angle OAF =A/2 - (90^(circ) -C)`

`=A/2 +C - (A+B+C)/2 =(C-B)/2`

Also , `AI =IE /(sin A/2) =r/(sin A/2) = (4 R sin (A/2) sin (B/2) sin (C/2))/(sin (A/2))`

so, `AI =4R sin B/2 sin C/2`

`:.` `OI^2 = OA^2 + Al^2 - 20A * AI cos (OAI)`

`OI^2= R^2 +16R^2 sin^2 B/2 sin^2 C/2 - 8 R^2 sin B/2 sin C/2 cos ((C-B)/2)`
`OI^2/R^2 =1 +16 sin^2 (B/2) sin^2 (C/2) -8 sin (B/2) sin (C/2) [ cos (B/2) cos (C/2) +sin (B/2) sin (C/2)]`

`= 1-8 sin (B/2) sin (C/2 ) [cos (B/2)cos (C/2) -sin (B/2) sin (C/2)]`

`= 1-8 sin (B/2) sin (C/2) cos ((B+C)/2) =1 -8 sin (A/2) sin(B/2) sin (C/2)`

`OI =R sqrt( 1-8 sin(A/2) sin (B/2) sin(C/2))`

we can write this in another form also.

OI^2=R^2 -8R^2 sin (A/2) sin (B/2) sin (C/2)`

`= R^2 -2R( 4R sin (A/2) sin(B/2) sin (C/2))`

`=R^2 -2Rr` as `r=4R sin A/2sin B/2 sin C/2`

Polygon :

(i) Sum of interior angles of a polygon `= (n - 2) x pi`. where `n ge 2` and `n` denotes number of sides of a
polygon.
(ii) Sum of exterior angles of a polygon is `2pi`.
(iii) Convex polygon : If the highest interior angle is less than `180^(circ)` then it is called convex polygon.
(iv) Concave polygon: Highest interior angle is more than `180^(circ)` then it is concave polygon.

Cyclic Quadrilateral :

A cyclic quadrilateral is a quadrilateral which can be inscribed by a circle.

Note: The sum of the opposite angles of a cyclic quadrilateral is `180^(circ)`
In a cyclic quadrilateral sum of the products of the opposite sides is equal to the product of the diagonals.
Regular Polygon
A regular polygon is a polygon which has all its sides as well as its angles equal. If the polygon has `n` sides
sum of its internal angles is `(n - 2)pi` and each angle is `((n-2) pi)/(n)`


Note : In the regular polygon, the centroid, the circumcentre and the in-centre are the same.
To find the perimeter `(P)` and Area`( A)` of a regular polygon inscribed in a circle of radius `R`.
Let `AB, BC` and `CD` be three successive sides of the polygon and `O` be the centre of both the incircle
and the circumcircle of the polygon.


`angle BOC= 2pi/n` so `angle BOL =1/2 (2pi/n) = pi/n`
If `a` be the side of the polygon, we have

`a = BC = 2BL = 2Rsin (angle BOL) = 2 R sin (pi/n)`

so , `R=a/2cosec (pi/n)`

Again `a = 2BL = 2 OL tan (angle BOL), OL= a/(2 tan (pi/n))= a/2 cot (pi/n) => r =a/2 cot (pi/n)`

where `R`: Radius of circle circumscribing the polygon `= OB = OC`
`r`: Radius of circle inscribed in the polygon `= OL`
Perimeter `P = nBC = n(2BL) = 2n R sin (angle BOL) = 2n R sin(pi/n)= 2 n R sin (pi/n)`

Area `A = n Area of Delta BOC= n 1/2 R*R* sin (angle BOC)=(nR^2)/2sin (2pi/n)`

When three elements of a triangle are known, the other three elements can be evaluated. This process is
called solution of triangles. Note following points

(i) If the three sides `a, b, c` are given, angle `A` is a obtained from
`tanA/2 = sqrt (((s-b)(s-c))/(s (s-a)))` or `cos A = (b^2+c^2 -a^2)/(2bc)`

`B` and `C` can be obtained similarly.

(ii) If two sides `b` and `c` and the included angle `A` are given, then

`tan((B-C)/2) = (b-c)/(b+c) cot A/2` gives `(B-C)/2 ` also ` (B+C)/2=90^(circ) -A/2`

so that `B` and `C` can be evaluated. The third side is given by
`a= (b sin A)/(sin B)` or `a^2 =b^2+c^2-2bc cos A`

(iii) If two sides `b` and `c` and the angle `B` (opposite to side `b`) are given, then `sin C=c/b sin B`

`A=180^(circ) - (B+C)` and `a=(b sin A)/(sin B)` give the remaining elements.

Here in this segment there are many cases of possibility of triangle. We will study them one by one.

Case-I :
`b < c sin B`,

We draw the side `c` and angle `B`. Such kind of triangle is not possible.

Case-II :
`b = c sin B` and `B` is an acute angle, then there is only one triangle possible.

Case-Ill :
If ` b > c sin B, b < c` and `B` is an acute angle, then there are two values of angle `C`.

Case-IV:
`b > c sin B, c < b` and `B` is an acute angle, then there is only one triangle possible.

Case-V :
`b > c sin B, c > b` and `B` is an obtuse angle. For any choice of point `C, b` will be greater
than `c` which is a contradiction as `c > b` (given). So there is no triangle possible.
Because `B` is obtuse .

Case-VI:
`b > c sin B, c < b` and `B` is an obtuse angle. We can see that the circle with `A` as
centre and `b` as radius will cut the line only in one point. So, one triangle is possible.

Case-VII:

`b > c` and `B = 90^(circ)`


Again the circle with `A` as centre and `b` as radius will cut the line only in one point. So
only one triangle is possible.

Case-VIII:
`b le c` and `B = 90^(circ)`

The circle with `A` as centre and `b` as radius will not cut the line in any point. So, no triangle is possible.
Point `C` will coincide with point `B`

Alternative method : By applying the cosine rule, we have

`cos B=(a^2+c^2-b^2)/2ac`

`=> a^2-(2ccos B) a +c^2 -b^2 =0`

`=> a= c cos B pm sqrt ([ (c cos B)^2-(c^2 -b^2)])`

`=> a = c cos B pm sqrt ( [ b^2 -(c sin B)^2])`

This equation leads to the following cases:

Case-I : lf `b < c sin B`, no such triangle is possible.

Case-II: Let `b = c sin B`. There are further following two cases:

(a) `B` is an obtuse angle `=> cos B` is negative. There exists no solution triangle.

(b) `B` is an acute angle `=> cos B` is positive. There exists only one such triangle.

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

(a) `B` is an acute angle `=> cos B` is positive. In this cases two values of `a` will exists if and only if

`c cos B > sqrt (b^2-(c sin B)^2) ` or `c >b` `=>` two such triangles are possible.
If `c < b`, only one such triangle is possible.


(b) `B` is an obtuse angle `=>` ` cos B` is negative. In this case, triangle will exists if and only if


`sqrt (b^2 -(c sin B)^2) > |cos B| => b >c`. So in this case only one triangle is possible.
If `b < c` there exists no such triangle.