# Solution OF Triangle

### Basic Rules :

In a triangle ABC, the angles are denoted by capital letters A,B, and C and the lengths of the sides opposite to these angles are denoted by small letters a, b, and c respectively.

Semi – perimeter of the triangle is written as  $\large s = \frac{a+b+c}{2}$ and its area denoted by S or Δ . Sine rule:

$\Large \frac{sinA}{a} = \frac{sinB}{b} = \frac{sinC}{c} = \frac{1}{2R}$

, where R is the radius of the circumcircle of the Δ ABC.

Cosine rule:

$\large cosA = \frac{b^2 + c^2 – a^2}{2 b c}$

$\large cosB = \frac{c^2 + c^2 – b^2}{2 a c}$

$\large cosC = \frac{a^2 + b^2 – c^2}{2 a b}$

Projection rule:

a = b cosC + c cosB ,

b = c cosA + a cosC ,

c = a cosB + b cosA

Napier’s analogy :

$\large tan\frac{B-C}{2} = \frac{b-c}{b+c}cot\frac{A}{2}$

$\large tan\frac{C-A}{2} = \frac{c-a}{c+a}cot\frac{B}{2}$

$\large tan\frac{A-B}{2} = \frac{a-b}{a+b}cot\frac{C}{2}$

##### Solutions of Triangle : m-n Theorem If in a triangle ABC, D is a point on the line BC such that BD:DC = m : n and ∠ADC = θ , ∠BAD = α , ∠DAC = β , then

(a) (m + n)cotθ = m cotα – n cotβ

(b) (m + n)cotθ = n cotB – m cotC

##### Auxiliary Formulae:

Trigonometric ratios of half – angles:

$\large sin\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{b c}}$

$\large sin\frac{B}{2} = \sqrt{\frac{(s-c)(s-a)}{c a}}$

$\large sin\frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{a b}}$

$\large cos\frac{A}{2} = \sqrt{\frac{s(s-a)}{b c}}$

$\large cos\frac{B}{2} = \sqrt{\frac{s(s-b)}{c a}}$

$\large cos\frac{C}{2} = \sqrt{\frac{s(s-c)}{a b}}$

$\large tan\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

$\large tan\frac{B}{2} = \sqrt{\frac{(s-c)(s-a)}{s(s-b)}}$

$\large tan\frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

##### Area of a triangle:

$\large \Delta = \frac{1}{2}b c sinA = \frac{1}{2}c a sinB = \frac{1}{2}a b sinC$

$\large \Delta = \sqrt{s(s-a)(s-b)(s-c)} = \frac{a b c}{4 R}$

where R and r are the radii of the circumcircle and the incircle of the Δ ABC respectively.

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