# Properties of Triangles – Tangent Rule and Area of Triangle

## Properties of Triangles – Tangent Rule and Area of Triangle

Tangent rule (or) Napier’s analogy:

In ΔABC,

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

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

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

Area of ΔABC is given by:

(I) Δ = ½ ab sin C; Δ = ½ bc sin A; Δ = ½ ca sin B,

(II) $$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$$,

(III) $$\Delta =\frac{abc}{4R}$$,

(IV) Δ = 2R² sin A sin B sin C.

(V) Δ = rs

(VI) $$\Delta =\sqrt{r{{r}_{1}}{{r}_{2}}{{r}_{3}}}$$.

Example: Prove that $$\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{abc}{{{\Delta }^{3}}}=\frac{4R}{{{r}^{2}}{{S}^{2}}}$$.

Solution: Given that,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{abc}{{{\Delta }^{3}}}=\frac{4R}{{{r}^{2}}{{S}^{2}}}$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)=\frac{S}{\Delta }-\frac{S-a}{\Delta }$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)=\frac{S-S+a}{\Delta }$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)=\frac{a}{\Delta }$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)=\frac{S}{\Delta }-\frac{S-b}{\Delta }$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)=\frac{S-S+b}{\Delta }$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)=\frac{b}{\Delta }$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{S}{\Delta }-\frac{S-c}{\Delta }$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{S-S+c}{\Delta }$$,

$$\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{c}{\Delta }$$,

L.H.S  $$\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)$$,

$$=\frac{a}{\Delta }\times \frac{b}{\Delta }\times \frac{c}{\Delta }$$,

$$=\frac{abc}{{{\Delta }^{3}}}$$,

$$=\frac{4R\times \Delta }{{{\Delta }^{3}}}$$,

$$=\frac{4R}{{{\Delta }^{2}}}$$,

$$=\frac{4R}{{{\left( rS \right)}^{2}}}$$ = R.H.S.

Hence proved $$\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{abc}{{{\Delta }^{3}}}=\frac{4R}{{{r}^{2}}{{S}^{2}}}$$.