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

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 

[latexpage]

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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