Inside Story
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1 PROPERTIES AND SOLUTIONS OF TRIANGLE
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1.1 POINTS TO BE REMEMBERED
- 1.1.1 LAWS OF SINE OR SINE RULE
- 1.1.2 LAWS OF COSINES OR COSINES RULE–
- 1.1.3 PROJECTION FORMULA –
- 1.1.4 LAWS OF TANGENT OR TANGENT RULES (Napier’s Analogy)
- 1.1.5 HALF ANGLE FORMULAE OR SEMI-SUM FORMULAE
- 1.1.6 AREA OF TRIANGLE –
- 1.1.7 HERON’S FORMULA –
- 1.1.8 CIRCUMCIRCLE OF A TRIANGLE
- 1.1.9 INCIRCLE OF A TRIANGLE
- 1.1.10 EXCRIBED CIRCLE OF A TRIANGLE
- 1.1.11 ORTHOCENTRE OF A TRIANGLE
- 1.1.12 CYCLIC QUADRILATERAL
- 1.1.13 PTOLEMY THEOREM –
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1.1 POINTS TO BE REMEMBERED
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]
Where k is some constant.
LAWS OF COSINES OR COSINES RULE–
\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
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 –
HERON’S FORMULA –
CIRCUMCIRCLE OF A TRIANGLE
In any triangle ABC,
INCIRCLE OF A TRIANGLE
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
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
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-
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= $$\frac{a}{2cot\left(\frac{\pi}{n}\right)}$$