Question

Draw the incircle of the $\Delta ABC$ with $AB=5$ cm, $\angle B={60}^o$ and $BC=7$ cm

Answer 1

R.E.F image

$AB=5cm$

$\angle B=60$

$BC=7cm$

by cosine Rule

$cosB=\frac{a^2+c^2-b^2}{2ac}$

$cos{60}^{\circ }=\frac{\left(5{)}^2\right(7{)}^2-b^2}{2-5-7}$

$\frac{1}{2}=\frac{74-b^2}{70-35}$

$b^2=74-35$

$b^2=39$

$b=\sqrt{39}$

In radius $\left(r\right)=\frac{2\times Areaof\Delta ABC}{sumofsides}$

In radius $\left(r\right)=\frac{2\times \frac{1}{2}\times ac.sinB}{(a+b+c)}$

In radius $\left(r\right)=\frac{5\times 7\times sin{60}^{\circ }}{(5+7+\sqrt{39})}=\frac{35\times \frac{\sqrt{3}}{2}}{12+\sqrt{39}}$

$=\frac{35\sqrt{3}}{2(12+\sqrt{39})}=\frac{35\sqrt{3}}{2(12+\sqrt{13\times 3})}$

divide by $\sqrt{3}$ in denometor

and Now

$=\frac{35}{2(4\sqrt{3}+\sqrt{13})}$

$=\frac{35(4\sqrt{3}-\sqrt{13})}{2(48-13)}=\frac{35(4\sqrt{3}-\sqrt{13})}{2\left(35\right)}$

$r=2\sqrt{3}-\frac{1}{2}\sqrt{13}$