Revision: Properties of Triangles JEE Main Properties of Triangles

A circle is defined as the figure (closed curve) obtained by joining all those points in a plane which are at a constant distance from a fixed point in the same plane. 

• Centre → Fixed point

• Radius → Constant distance

• Circumference → Perimeter of the circle

The straight line joining the eye of the observer to the point on the object being viewed.

The angle between the line of sight and the horizontal through the observer’s eye, when the object is below the level of the observer’s eye.

The angle between the line of sight and the horizontal through the observer’s eye, when the object is above the level of the observer’s eye.

\[\mathrm{In~\Delta ABC,~\frac{a}{\sin A}=\frac{b}{sinB}=\frac{c}{sinC}}\]

In ΔABC,

i.  \(\mathrm{cos}A=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2\mathrm{b}\mathrm{c}}\)

ii. \[\mathrm{cos}\mathrm{B}=\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2\mathrm{c}\mathrm{a}}\]

iii. \[\mathrm{cos}\mathrm{C}=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}}{2\mathrm{ab}}\]

• x = r cosθ
• y = r sinθ
• \[\tan\theta=\frac{y}{x}\]
• \[\mathbf{r}=\sqrt{x^2+y^2}\]

In ΔABC,

In ΔABC, if a + b + c = 2s, then

1. \[\sin\frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s-b})(\mathrm{s-c})}{\mathrm{bc}}}\]

\[\sin\frac{\mathrm{B}}{2}=\sqrt{\frac{(\mathrm{s-c})(\mathrm{s-a})}{\mathrm{ca}}}\]

\[\sin\frac{\mathrm{C}}{2}=\sqrt{\frac{(\mathrm{s-a})(\mathrm{s-b})}{\mathrm{ab}}}\]

2. \[\cos\frac{\mathrm{A}}{2}=\sqrt{\frac{\mathrm{s(s-a)}}{\mathrm{bc}}}\]

\[\cos\frac{\mathrm{B}}{2}=\sqrt{\frac{\mathrm{s(s-b)}}{\mathrm{ca}}}\]

\[\cos\frac{\mathrm{C}}{2}=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}}\]

3. \[\tan\frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s-b})(\mathrm{s-c})}{\mathrm{s(s-a)}}}\]

\[\tan\frac{\mathrm{B}}{2}=\sqrt{\frac{(\mathrm{s-c})(\mathrm{s-a})}{\mathrm{s(s-b)}}}\]

\[\tan\frac{\mathrm{C}}{2}=\sqrt{\frac{(\mathrm{s-a})(\mathrm{s-b})}{\mathrm{s(s-c)}}}\]

Area of ΔABC = \[\frac{1}{2}\mathrm{ab~sinC}\]

                         = \[=\frac{1}{2}\mathrm{bc~sinA}=\frac{1}{2}\mathrm{ac~sinB}\]

Heron’s Formula:

The area of ΔABC = \[\sqrt{\mathrm{s(s-a)(s-b)(s-c)}}\] 

where, 2s = a + b + c

In ΔABC,

i. \[\tan\left(\frac{\mathrm{A-B}}{2}\right)=\left(\frac{\mathrm{a-b}}{\mathrm{a+b}}\right)\cot\frac{\mathrm{C}}{2}\]

ii. \[\tan\left(\frac{\mathrm{B-C}}{2}\right)=\left(\frac{\mathrm{b-c}}{\mathrm{b+c}}\right)\cot\frac{\mathrm{A}}{2}\]

iii. \[\tan\left(\frac{\mathrm{C-A}}{2}\right)=\left(\frac{\mathrm{c-a}}{\mathrm{c+a}}\right)\cot\frac{\mathrm{B}}{2}\]

By sine rule, `a/(sin A) = b/(sin B) = c/(sin C) = k`

∴ a = k sin A, b = k sin B, c = k sin C

RHS = `((a - b)/(a + b)) cot (C/2)`

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

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

= `(2 cos ((A + B)/2)*sin((A - B)/2))/(2 sin ((A + B)/2)*cos((A - B)/2)) xx (cos(C/2))/(sin(C/2))`

= `(cos(pi/2 - C/2)*sin((A - B)/2))/(sin(pi/2 - C/2)*cos((A - B)/2)) xx (cos (C/2))/(sin(C/2))`     ...[∵A + B + C = π]

= `(sin(C/2))/(cos(C/2)) xx tan ((A - B)/2) xx (cos (C/2))/(sin(C/2))`

= `tan ((A - B)/2)` = LHS

• Concentric circles → Same centre, different radii

• Equal circles → Same radius

• Circumscribed circle → Circle passes through all vertices of a polygon
  Centre → Circumcentre

• Inscribed circle → Circle touches all sides of a polygon
  Centre → Incentre