Properties and solutions of triangle for IIT-JEE (mains & advanced) | chapter notes | study material class 12.

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PROPERTIES AND SOLUTIONS OF TRIANGLE

Here, there are some key concepts regarding the properties and solutions of triangles. these are very important for the students who preparing for IIT-JEE and other equivalent exams.

POINTS TO BE REMEMBERED

 LAWS OF SINE OR SINE RULE 

The sides of a triangle are proportional to the sines of opposite angles i.e, In an ∆ABC,  

    \[\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = k\]

Where k is some constant.

LAWS OF COSINES OR COSINES RULE

   In any ∆ABC , We have

    \begin{align*} CosA& = \frac{\left(b^2 +c^2 - a^2\right)}{2bc}\\ CosB& = \frac{\left(c^2 + a^2 - b^2\right)}{2ac}\\ CosC &= \frac{\left( a^2 + b^2 - c^2\right)}{2ab} \end{align*}

PROJECTION FORMULA –

In any ∆ABC

    \begin{align*} a& = b.cosC + c.cosB\\ b& = a.cosC + c.cosA\\ c& = a.cosB + b.cosA \end{align*}

LAWS OF TANGENT OR TANGENT RULES (Napier’s Analogy)

    \begin{align*} tan\left(\frac{B-C}{2}\right)&= \left(\frac{b-c}{b+c}\right)cot\left(\frac{A}{2}\right)\\ tan\left(\frac{A-B}{2}\right)& = \left(\frac{a-b}{a+b}\right)cot\left(\frac{C}{2}\right)\\ tan\left(\frac{C-A}{2}\right)& = \left(\frac{c-a}{c+a}\right)cot\left(\frac{B}{2}\right) \end{align*}

HALF ANGLE FORMULAE OR SEMI-SUM FORMULAE

    \begin{align*} sin\left(\frac{A}{2}\right)& = \sqrt{\frac{(s-b)(s-c)}{bc}}\\ sin\left(\frac{B}{2}\right)& = \sqrt{\frac{(s-c)(s-a)}{ac}}\\ sin\left(\frac{C}{2}\right)& = \sqrt{\frac{(s-a)(s-b)}{ab}} \end{align*}

    \begin{align*} cos\left(\frac{A}{2}\right)&= \sqrt{\frac{s(s-a)}{bc}}\\ cos\left(\frac{B}{2}\right)& = \sqrt{\frac{s(s-b)}{ac}}\\ cos\left(\frac{C}{2}\right)& = \sqrt{\frac{s(s-c)}{ab}} \end{align*}

    \begin{align*} tan\left(\frac{A}{2}\right)& = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\\ tan\left(\frac{B}{2}\right)& = \sqrt{\frac{(s-c)(s-a)}{s(s-b}}\\ tan\left(\frac{C}{2}\right)& = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \end{align*}

AREA OF TRIANGLE

The area of any triangle ABC can be given by:

    \[\Delta = \frac{bc.sinA}{2} = \frac{ac.sinB}{2} = \frac{ab.sinC}{2}\]

HERON’S FORMULA –

In a ∆ABC , if a+b+c = 2s, then it’s area can given by:

    \[\Delta = \sqrt{s(s-a)(s-b)(s-c)}\]

CIRCUMCIRCLE OF A TRIANGLE

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

In any triangle ABC,

a).

    \[R = \frac{a}{2sinA}= \frac{b}{2sinB} =\frac{ c}{2sinC}\]

b).

    \[R = \frac{abc}{4\Delta}\]

Everywhere  means area

INCIRCLE OF A TRIANGLE

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

In any triangle ABC,
a).

    \[r = \frac{\Delta}{s}\]

b).

    \[r = (s-a)tan\left(\frac{A}{2}\right)= (s-b)tan\left(\frac{B}{2}\right)= (s-c)tan\left(\frac{C}{2}\right)\]

c).

    \begin{align*} r& = \frac{a.sin\left(\frac{B}{2}\right).sin\left(\frac{C}{2}\right)}{cos\left(\frac{A}{2}\right)}\\ r &= \frac{b.sin\left(\frac{A}{2}\right).sin\left(\frac{C}{2}\right)}{cos\left(\frac{B}{2}\right)}\\ r& = \frac{c.sin\left(\frac{B}{2}\right).sin\left(\frac{A}{2}\right)}{cos\left(\frac{C}{2}\right)} \end{align*}

d).

    \[r= 4Rsin\left(\frac{A}{2}\right).sin\left(\frac{B}{2}\right).sin\left(\frac{C}{2}\right)\]

EXCRIBED CIRCLE OF A TRIANGLE

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

In any triangle ABC
a).

    \begin{align*} r_1& =\frac{\Delta}{s-a}\\ r_2&= \frac{\Delta}{s-b}\\ r_3&= \frac{\Delta}{s-c} \end{align*}

b).

    \begin{align*} r1& = s.tan\left(\frac{A}{2}\right)\\ r2& = s.tan\left(\frac{B}{2}\right)\\ r3& = s.tan\left(\frac{C}{2}\right) \end{align*}

c).

    \begin{align*} r_1 = \frac{a.cos\left(\frac{B}{2}\right).cos\left(\frac{C}{2}\right)}{cos\left(\frac{A}{2}\right)}\\ r_2 = \frac{b.cos\left(\frac{C}{2}\right).cos\left(\frac{A}{2}\right)}{cos\left(\frac{B}{2}\right)}\\ r_3 = \frac{c.cos\left(\frac{A}{2}\right).cos\left(\frac{B}{2}\right)}{cos\left(\frac{C}{2}\right)} \end{align*}

d).

    \begin{align*} r_1& = 4R.sin\left(\frac{A}{2}\right).cos\left(\frac{B}{2}\right).cos\left(\frac{C}{2}\right)\\ r_2&= 4R.sin\left(\frac{B}{2}\right).cos\left(\frac{A}{2}\right).cos\left(\frac{C}{2}\right)\\ r_3&= 4R.sin\left(\frac{C}{2}\right).cos\left(\frac{A}{2}\right).cos\left(\frac{B}{2}\right) \end{align*}

ORTHOCENTRE OF A TRIANGLE

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

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

2RcosBcosC , 2RcosCcosA , 2RcosAcosB

POINTS TO BE NOTED

1). The circumradius of a padel triangle is 

    \[\frac{R}{2}\]

2). Area of a pedal triangle is

    \[2\Delta cosA.cosB.cosC = \frac{R^2.sin2A.sin2B.sin2C}{2}\]

3). Length of the median AD, BE, and CF of the ∆ABC. is given by-

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

AD =

    \[\frac{\sqrt{(b^2+c^2+2bc.cosA)}}{2} =\frac{\sqrt{(2b^2+2c^2-a^2)}}{2}\]

BE =

    \[\frac{\sqrt{(a^2+c^2+2ac.cosB)}}{2} = \frac{\sqrt{(2a^2+2c^2-b^2)}}{2}\]

CF =

    \[\frac{\sqrt{(b^2+a^2+2ba.cosC)}}{2} =\frac{\sqrt{(2b^2+2a^2-c^2)}}{2}\]

4). Length of the angle bisector through vertex A=

    \[\frac{2bc.cos\left(\frac{A}{2}\right)}{(b+c)}\]

 Through vertex B =

    \[\frac{2ac.cos\left(\frac{B}{2}\right)}{(a+c)}\]

 Through vertex C =

    \[\frac{2ab.cos\left(\frac{C}{2}\right)}{(a+b)}\]

5). Length of the altitude from;

  From vertex A =

    \[\frac{a}{cotB+cotC}\]

  From vertex B =

    \[\frac{b}{cotC+cotA}\]

  From vertex C =

    \[\frac{c}{cotA+cotB}\]

6). Distance between circumcentre and the orthocentre =

    \[R\sqrt{(1-8cosA.cosB.cosC)}\]

7). Distance between circumcentre and incentre is=

    \[\sqrt{(R^2-2Rr)}=R\sqrt{\left(1-8cos\left(\frac{A}{2}\right).cos\left(\frac{B}{2}\right).cos\left(\frac{C}{2}\right)\right)}\]

8). Distance between incentre and the orthocentre =

    \[\sqrt{(2r^2-4R^2.cosA.cosB.cosC)}\]

9). Distance between circumcentre and centroid =

    \[R^2-\frac{(a^2+b^2+c^2)}{9}\]

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 center of the escribed circles opposite to the angle A, B, and C respectively.  And O is the orthocentre. Then

    \[OI_1 =R\sqrt{\left(1+8sin\left(\frac{A}{2}\right).cos\left(\frac{B}{2}\right).cos\left(\frac{C}{2}\right)\right)}\]

    \[OI_2 = R\sqrt{\left(1+8cos\left(\frac{A}{2}\right).sin\left(\frac{B}{2}\right).cos\left(\frac{C}{2}\right)\right)}\]

    \[OI_3 = R\sqrt{\left(1+8cos\left(\frac{A}{2}\right).cos\left(\frac{B}{2}\right).sin\left(\frac{C}{2}\right)\right)}\]

CYCLIC QUADRILATERAL

POINTS TO BE NOTED

1). Area of the cyclic quadrilateral ABCD is =

    \[\sqrt{s(s-a)(s-b)(s-c)(s-d)}\]

2). Circumradius of the cyclic quadrilateral =

   

    \[\sqrt{\frac{(ab+cd)(ad+bc)(ac+bd)}{16(s-a)(s-b)(s-c)(s-d)}}\]

3). CosB = \frac{(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 the sum of its internal angle is (n-2)π and each angle is

    \[\frac{\pi(n-2)}{n}\]

2). In a regular polygon the centroid, the circumcentre, and incentre are the same.

3). Area of the regular polygon =

    \[\frac{na^2.cot\left(\frac{\pi}{n}\right)}{4}\]

 

    \[=\frac{ nR^2.sin\left(\frac{2\pi}{n}\right)}{2} = nr^2.tan\left(\frac{\pi}{n}\right)\]

Where n is the no of sides of the regular polygon, R is the radius of circumscribing circle, and r is the radius of the incircle of a polygon.

4). Radius of circumscribing circle =

    \[R = \frac{a}{2sin\left(\frac{\pi}{n}\right)}= \frac{a}{2csc\left(\frac{\pi}{n}\right)}\]

Where a is the length of the side of the regular polygon of n sides.

5). Radius of inscribed circles r=

    \[\frac{a}{2cot\left(\frac{\pi}{n}\right)}\]

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