NIMCET 2019 — Mathematics PYQ

Concept:
Let a b and c are three sides of a triangle
$s=\frac{a+b+c}{2}$, $\Delta =s(s-a)\tan \frac{A}{2}$
$\tan A=\frac{2\tan \frac{A}{2}}{1-{\tan }^2\frac{A}{2}}$

Calculations:

Given, $\Delta ={\mathrm{a}}^2-($b-c${)}^2$, where $\Delta$ is the area of the $\Delta$ABC.

$$\Rightarrow \Delta =[\begin{array}{c}\mathrm{a}-(\mathrm{b}-\mathrm{c})[\mathrm{a}-(\mathrm{b}-\mathrm{c})]\end{array}$$

$$\Rightarrow \Delta =(\mathrm{a}-\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}-\mathrm{c})$$

We know that s$=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$

$$\Rightarrow \Delta =4(s-c)(s-b)$$

Again, we know that $\Delta =s\left(s-a\right)\tan \frac{\mathrm{A}}{2}$

$\Rightarrow s\left(s-\mathrm{a}\right)\tan \frac{\mathrm{A}}{2}=4\left(\mathrm{s}-\mathrm{c}\right)($s- b) $\Rightarrow \tan \frac{\mathrm{A}}{2}=4$ $\frac{\left(\mathrm{s}-\mathrm{c}\right)\left(\mathrm{s}-\mathrm{b}\right)}{\mathrm{s}(\mathrm{s}-\mathrm{a})}$

$\Rightarrow \tan \frac{\mathrm{A}}{2}=4{\tan }^2\frac{\mathrm{A}}{2}$

$\Rightarrow \tan \frac{\mathrm{A}}{2}=\frac{1}{4}$

$\mathbf{We}\mathbf{know}\mathbf{that},\mathbf{tan}\mathbf{A}=\frac{2\tan \frac{\mathrm{A}}{2}}{1-{\tan }^2\frac{\mathrm{A}}{2}}$