” PROPERTIES AND SOLUTIONS OF TRIANGLE for IIT-JEE (mains & advanced), AIEEE , NTSE , KVPY , and others engineering entrance examination.

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PROPERTIES AND SOLUTIONS OF TRIANGLE

** POINTS TO BE REMEMBERED
 
*  LAWS OF SINE OR SINE RULE-
    The sides of a triangle are proportional to the sines of the opposite angles. That is , In a ∆ABC, we have;
                   a/sinA = b/sinB = c/sinC = k
Where k is some constant.
* LAWS OF COSINES OR COSINES RULE
   In any ∆ABC , We have;
 a). CosA = (b^2 +c^2 – a^2)/2bc
 b). CosB = (c^2 + a^2 – b^2)/2ac
 c). CosC = ( a^2 + b^2 – c^2)/2ab
* PROJECTION FORMULA – In any ∆ABC
a). a = bCosC + cCosB
b). b = aCosC + cCosA
c). c = aCosB + bCosA
* LAWS OF TANGENT OR TANGENT RULES (Napier’s Analogy)
a). tan(B-C/2) = (b-c/b+c).cot(A/2)
b). tan(A-B/2) = (a-b/a+b).cot(C/2)
c). tan(C-A/2) = (c-a/c+a).cot(B/2)
* HALF ANGLE FORMULAE OR SEMI-SUM FORMULAE
a). sin(A/2) = √(s-b)(s-c)/bc ,
      sin(B/2) = √(s-c)(s-a)/ac
      sin(C/2) = √(s-a)(s-b)/ab
b). cos(A/2) = √s(s-a)/bc
      cos(B/2) = √s(s-b)/ac
      cos(C/2) = √s(s-c)/ab
c). tan(A/2) = √(s-b)(s-c)/s(s-a)
     tan(B/2) = √(s-c)(s-a)/s(s-b
     tan(C/2) = √(s-a)(s-b)/s(s-c)
* THE AREA OF TRIANGLE – The area of any triangle ABC can be given by :
∆ = (1/2)bc.sinA = (1/2)ac.sinB = (1/2)ab.sinC
* HERON’S FORMULA – In a ∆ABC , if a+b+c = 2s, then it’s area can given by:
         ∆ = √s(s-a)(s-b)(s-c)
* CIRCUMCIRCLE OF A TRIANGLE

In any triangle ABC,

a). R = a/2sinA = b/2sinB = c/2sinC
b). R = abc/4∆.    [Everywhere ∆ means area]
* INCIRCLE OF A TRIANGLE

In any triangle ABC,

a). r = ∆/s
b). r = (s-a)tan(A/2)= (s-b)tan(B/2)= (s-c)tanC/2
c). r = a.sin(B/2).sin(C/2)/cos(A/2)
     r = b.sin(A/2).sin(C/2)/cos(B/2)
     r = c.sin(B/2).sin(A/2)/cos(C/2)
d). r= 4Rsin(A/2).sin(B/2).sin(C/2)
 
* EXCRIBED CIRCLE OF A TRIANGLE
In any triangle ABC
a). r1 = ∆/s-a
      r2 = ∆/s-b
      r3= ∆/s-c
b). r1 = s.tan(A/2)
      r2 = s.tan(B/2)
      r3 = s.tan(C/2)
c). r1 = a.cos(B/2).cos(C/2)/cos(A/2)
     r2 = b.cos(C/2).cos(A/2)/cos(B/2)
     r3 = c.cos(A/2).cos(B/2)/cos(C/2)
d). r1 = 4R.sin(A/2).cos(B/2).cos(C/2)
     r2= 4R.sin(B/2).cos(A/2).cos(C/2)
     r3= 4R.sin(C/2).cos(A/2).cos(B/2)
 
* ORTHOCENTRE OF A TRIANGLE

The triangle formed by the joining the points F , E and D is called pedal triangle.

In any triangle ABC
a). The distance of the orthocentre from the vertices of the triangle ABC is;
Distance from A = 2RcosA
Distance from B = 2RcosB
Distance from C = 2RcosC
b). Distance of the orthocentre from the sides of the triangle ABC is;
2RcosB.cosC , 2RcosC.cosA , 2RcosA.cosB
 
** POINTS TO BE NOTED
1). Circumradius of a padel triangle is = R/2
2). Area of a pedal triangle is = 2∆cosA.cosB.cosC = R^2.sin2A.sin2B.sin2C/2
3). Length of the median AD , BE and CF of the ∆ABC. is given by;

* AD = (1/2)√(b^2+c^2+2bc.cosA =(1/2)√(2b^2+2c^2-a^2)
* BE = 1/2)√(a^2+c^2+2ac.cosB =(1/2)√(2a^2+2c^2-b^2)
* CF = 1/2)√(b^2+a^2+2ba.cosC =(1/2)√(2b^2+2a^2-c^2)
4). Length of the angle bisector through vertex A= 2bc.cos(A/2)/b+c
 * Through vertex B = 2ac.cos(B/2)/a+c
 * Through vertex C = 2ab.cos(C/2)/a+b
5). Length of the altitude from;
  * From vertex A = a/cotB+cotC
  * From vertex B = b/cotC+cotA
  * From vertex C = c/cotA+cotB
6). Distance between circumcentre and the orthocentre = R√(1-8cosA.cosB.cosC)
7). Distance between circumcentre and incentre is=√(R^2-2Rr)=R√(1-8cos(A/2).cos(B/2).cos(C/2)
8). Distance between incentre and the orthocentre = √(2r^2-4R^2.cosA.cosB.cosC)
9). Distance between circumcentre and centroid =R^2-(a^2+b^2+c^2)/9
10). If circumcentre , centroid and orthocentre are collinear and G divides OO’ in the ratio of 1:2
11). If I1 , I2 and I3 are the centre of the excribed circles opposite to the angle A , B and C respectively.  And O is the orthocentre. Then
* OI1 = R√(1+8sin(A/2).cos(B/2).cos(C/2))
* OI2 = R√(1+8cos(A/2).sin(B/2).cos(C/2))
* OI3 =R√(1+8cos(A/2).cos(B/2).sin(C/2))
* CYCLIC QUADRILATERAL
 
** POINTS TO BE NOTED
 
1). Area of the cyclic quadrilateral ABCD is =
      √s(s-a)(s-b)(s-c)(s-d)
2). Circumradius of the cyclic quadrilateral =
   √[(ab+cd)(ad+bc)(ac+bd)/16(s-a)(s-b)(s-c)(s-d)]
3). CosB = (a^2+b^2-c^2-d^2)/2(ab+cd) and similarly other angles.
* PTOLEMY THEOREM – If ABCD is a cyclic quadrilateral then AC.BD = AB.CD+BC.AD
* REGULAR POLYGON
** POINTS TO BE NOTED
1). If a regular polygon has n sides , then sum of its internal angle is (n-2)π and each angle is
   (n-2)π /n
2). In a regular polygon the centroid , the circumcentre and incentre are same.
3). Area of the regular polygon = na^2.cot(π/n)/4
  = nR^2.sin(2π/n)/2 = nr^2.tan(π/n)
Where n is the no. of sides of the regular polygon. R is the radius of circumscribing circle, and r is the radius of INCIRCLE of polygon.
4). Radius of circumscribing circle =
    R = a/2sin(π/n) = a/2csc(π/n)
Where a is the length of the side of the regular polygon of n sides.
5). Radius of inscribed circles r= a/2cot(π/n)
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Suchit prajapati

A person who explores the whole universe just by sitting in his room.

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